Equivalent mass of ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ in its reaction with ${{\rm{I}}_2}$ is equal to :

$Molar{\rm{ }}mass{\rm{ }}/{\rm{ }}2$

$Molar{\rm{ }}mass{\rm{ }}/{\rm{ }}3$

$Molar{\rm{ }}mass{\rm{ }}/{\rm{ }}4$

SOME BASIC CONCEPTS OF CHEMISTRY MCQs for NEET UG CHEMISTRY — Easy, Page 4

91 Easy

91 of 114Question permalink

Understanding +4 | -1

Millimole of a solute in a solution can be given by :

View answer and explanation

Correct option: D

Both (b) and (c)

Option D is correct because millimoles can be calculated using both

M \times V_{in mL}

and

\frac{wt.}{mol.wt.} \times 1000

.

Solution approach

Correct Answer + Core Reason: Both (b) and (c). Millimoles can be calculated using both formulas.

Step-by-Step Solution:

Step 1: Define millimoles. Millimoles = Moles * 1000.

Step 2: Moles = Molarity (M) * Volume (in Liters).\nTherefore, Millimoles = M * V (in Liters) * 1000 = M * V (in mL).

Step 3: Moles = (Weight) / (Molecular weight).\nTherefore, Millimoles = (Weight / Molecular weight) * 1000.

Step 4: Final Answer: Both (b) and (c).

Why other options are wrong:
- Option A: $M \times {V_{{\rm{{in litre}}}}}$ gives moles, not millimoles.
- Option B: $M \times {V_{{\rm{{in mL}}}}}$ gives millimoles, but it's not the only correct option.
- Option C: $\frac{{{\rm{{wt}}\mathrm{.}}}}}{{{\rm{{mol}}\mathrm{.wt}\mathrm{.}}}} \times 1000$ gives millimoles, but it's not the only correct option.

Key Concepts:
- Moles and millimoles
- Molarity
- Molecular weight

Memory Aids:
- Remember the conversion between moles and millimoles.

Quick Tips:
- Pay attention to the units of volume.

92 Easy

92 of 114Question permalink

Understanding +4 | -1

In the relationship molecular formula = empirical formula

\times \;n.\;\mathrm{The}\;'n'\;\;\mathrm{may}\;\mathrm{ have}:

View answer and explanation

Correct option: C

Only positive integer value

Option C (Only positive integer value) is correct because 'n' in the formula must be a positive integer, representing the multiple of the empirical formula that makes up the molecular formula.

Solution approach

Correct Answer + Core Reason: Only positive integer value. The 'n' value must be a positive integer.

Step-by-Step Solution:

Step 1: Understand the relationship between molecular and empirical formulas.\nThe molecular formula represents the actual number of atoms of each element in a molecule, while the empirical formula represents the simplest whole-number ratio of atoms.

Step 2: Define 'n'.
'n' is the factor by which the subscripts in the empirical formula must be multiplied to obtain the molecular formula.

Step 3: Determine the possible values of 'n'.\nSince the molecular formula must have whole numbers of atoms, 'n' must be a positive integer (1, 2, 3, ...).

Step 4: Conclusion.\nTherefore, 'n' can only have a positive integer value.

Why other options are wrong:
- Option A: Any value is incorrect because 'n' must be a whole number.
- Option B: Zero value is incorrect because if n=0, the molecular formula would be nonexistent.
- Option D: None of the above is incorrect because option C is correct.

Key Concepts:
- Molecular formula
- Empirical formula

Memory Aids:
- Remember that the molecular formula is a whole-number multiple of the empirical formula.

Quick Tips:
- Calculate 'n' by dividing the molecular weight by the empirical formula weight.

93 Easy

93 of 114Question permalink

Understanding +4 | -1

On dissolving 1 mole each of the following acids in 1 litre water, the acid which do not give a solution of 1 N strength is:

View answer and explanation

Correct option: D

{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}

Option D (

{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}

) is correct because H3PO4 is a triprotic acid, meaning 1 mole of it will not produce a 1 N solution.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}$ . This is a triprotic acid, so 1 mole does not give 1 N solution.

Step-by-Step Solution:

Step 1: Define normality and molarity.

Step 2: Normality (N) = Molarity (M) * Basicity (for acids) or Acidity (for bases).

Step 3: HCl, HClO4, and HNO3 are monoprotic acids (basicity = 1). Therefore, 1 M solution = 1 N solution.

Step 4: H3PO4 is a triprotic acid (basicity = 3). Therefore, 1 M solution = 3 N solution.

Step 5: Thus, dissolving 1 mole of H3PO4 in 1 litre of water will give a 3 N solution, not a 1 N solution.

Why other options are wrong:
- Option A: HCl is a monoprotic acid, so 1 mole in 1 litre gives a 1 N solution.
- Option B: $\mathrm{HCl}{\mathrm{O}_4}$ is a monoprotic acid, so 1 mole in 1 litre gives a 1 N solution.
- Option C: $\mathrm{HN}{\mathrm{O}_3}$ is a monoprotic acid, so 1 mole in 1 litre gives a 1 N solution.

Key Concepts:
- Normality
- Molarity
- Basicity of acids
- Monoprotic, diprotic, and triprotic acids

Memory Aids:
- Normality = Molarity * n-factor (acidity/basicity).

Quick Tips:
- Always consider the basicity/acidity of the acid/base when converting between molarity and normality.

94 Easy

94 of 114Question permalink

Remembering +4 | -1

The net charge on ferrous ion is:

View answer and explanation

Correct option: A

+ 2

Option A (+2) is correct because the ferrous ion (Fe2+) has a charge of +2.

Solution approach

Correct Answer + Core Reason: +2. Ferrous ion is Fe $^{2+}$ .

Step-by-Step Solution:

Step 1: Recall the common oxidation states of iron.

Step 2: Iron can exist in two common oxidation states: +2 and +3.

Step 3: The ferrous ion is represented as Fe $^{2+}$ , indicating a +2 charge.

Step 4: The ferric ion is represented as Fe $^{3+}$ , indicating a +3 charge.

Why other options are wrong:
- Option B: +3 is the charge of the ferric ion (Fe $^{3+}$ ), not the ferrous ion.
- Option C: +4 is not a common oxidation state for iron.
- Option D: +5 is not a common oxidation state for iron.

Key Concepts:
- Oxidation states of transition metals
- Ferrous and ferric ions

Memory Aids:
- Ferrous = +2, Ferric = +3 (alphabetical order helps remember the increasing charge).

Quick Tips:
- Memorize common oxidation states of common ions.

95 Easy

95 of 114Question permalink

Application +4 | -1

Two elements

X

(atomic weight = 75) and

Y

(atomic weight =16) combine to give a compound having 75.8% of

X

. The formula of the compound is:

View answer and explanation

Correct option: D

{X_2}{Y_3}

Option D (

{X_2}{Y_3}

) is correct because the ratio of moles of X and Y calculated from the given percentage and atomic weights is approximately 2:3.

Solution approach

Correct Answer + Core Reason: ${X_2}{Y_3}$ . The formula is determined by finding the mole ratio of X and Y.

Step-by-Step Solution:

Step 1: Given: Atomic weight of X = 75, Atomic weight of Y = 16, Percentage of X = 75.8%

Step 2: Calculate the percentage of Y: 100% - 75.8% = 24.2%

Step 3: Calculate the moles of X: $\frac{{75.8}}{{75}} = 1.01$

Step 4: Calculate the moles of Y: $\frac{{24.2}}{{16}} = 1.51$

Step 5: Find the mole ratio of X to Y: $\frac{{1.01}}{{1.51}} \approx \frac{{2}}{{3}}$

Step 6: Therefore, the formula of the compound is ${X_2}{Y_3}$ .

Why other options are wrong:
- Option A: XY would imply a 1:1 mole ratio, which is incorrect based on the given percentages and atomic weights.
- Option B: ${X_2}Y$ would imply a 2:1 mole ratio, which is incorrect based on the given percentages and atomic weights.
- Option C: ${X_2}{Y_2}$ would imply a 1:1 mole ratio, which is incorrect based on the given percentages and atomic weights.
- Option D: This is the correct answer.

Key Concepts:
- Mole concept
- Empirical formula determination
- Percentage composition

Memory Aids:
- Remember that moles = (mass / atomic weight).

Quick Tips:
- Always convert percentages to masses (assuming 100g total mass) before calculating moles.

96 Easy

96 of 114Question permalink

Understanding +4 | -1

Number of g-atoms of an element in one atom are:

View answer and explanation

Correct option: B

1.66 \times {10^{ - 24}}

Option B (

1.66 \times {10^{ - 24}}

) is correct because one gram-atom is equal to Avogadro's number of atoms, and the mass of one atom in grams is the reciprocal of Avogadro's number.

Solution approach

Correct Answer + Core Reason: $1.66 \times {10^{ - 24}}$ . This is the mass of one atom in grams.

Step-by-Step Solution:

Step 1: Define gram-atom.

Step 2: One gram-atom of an element is the amount of the element that contains Avogadro's number ( $N_A = 6.022 \times 10^{23}$ ) of atoms and whose mass in grams is numerically equal to its atomic weight.

Step 3: The mass of one atom in grams is given by $\frac{{1}}{{{{N_A}}}} = \frac{{1}}{{{{6.022 \times 10^{23}}}}} = 1.66 \times 10^{-24}$ g.

Why other options are wrong:
- Option A: $6.023 \times {10^{23}}$ is Avogadro's number, which is the number of atoms in one gram-atom, not the mass of one atom in grams.
- Option C: $2 \times {10^{23}}$ is an arbitrary number and not related to the mass of one atom in grams.
- Option D: None of these is incorrect because option B is correct.

Key Concepts:
- Gram-atom
- Avogadro's number
- Atomic mass

Memory Aids:
- Mass of one atom = 1/Avogadro's number.

Quick Tips:
- Remember the relationship between gram-atom, Avogadro's number, and atomic mass.

97 Easy

97 of 114Question permalink

Analyzing +4 | -1

Moon takes 27.3 days to complete one orbit around the Earth. Now read the following statements and choose the correct code. Here T is for true statement and F is for ‘False statement’.
(i) Moon takes 655.2 hours to complete one orbit around the Earth.
(ii) Moon takes 39312 seconds to complete one orbit around the earth.
(iii) Moon takes 1638 minutes to complete one orbit around the Earth.

View answer and explanation

Correct option: C

T F F

Option C (T F F) is correct because statement (i) is true, while statements (ii) and (iii) are false based on the conversion of days to hours, seconds, and minutes.

Solution approach

Correct Answer + Core Reason: T F F. The statements are evaluated based on the conversion of days to hours, seconds, and minutes.

Step-by-Step Solution:

Step 1: Convert 27.3 days to hours:
27.3 days × 24 hours/day = 655.2 hours\nStatement (i) is true (T).

Step 2: Convert 27.3 days to seconds:
27.3 days × 24 hours/day × 60 minutes/hour × 60 seconds/minute = 2358720 seconds\nStatement (ii) is false (F) because 39312 seconds is incorrect.

Step 3: Convert 27.3 days to minutes:
27.3 days × 24 hours/day × 60 minutes/hour = 39312 minutes\nStatement (iii) is false (F) because 1638 minutes is incorrect.

Step 4: Conclusion:\nThe correct code is T F F.

Why other options are wrong:
- Option A (F T F): This is incorrect because statement (i) is true, not false.
- Option B (T T T): This is incorrect because statements (ii) and (iii) are false.
- Option D (T F T): This is incorrect because statement (iii) is false.

Key Concepts:
- Unit Conversion
- Time Conversion (days, hours, minutes, seconds)

Memory Aids:
- 1 day = 24 hours
- 1 hour = 60 minutes
- 1 minute = 60 seconds

Quick Tips:
- Always pay attention to units when performing conversions.
- Double-check your calculations to avoid errors.

98 Easy

98 of 114Question permalink

Understanding +4 | -1

The oxidation of

S{O_2}

to

S{O_3}

is an exothermic reaction. The yield of

S{O_3}

will be maximum if:

View answer and explanation

Correct option: B

Temperature is reduced and pressure is increased

The correct answer is B (Temperature is reduced and pressure is increased) because the reaction is exothermic and involves a decrease in the number of moles of gas, favoring low temperature and high pressure.

Solution approach

Correct Answer + Core Reason: Temperature is reduced and pressure is increased. The reaction is exothermic and involves a decrease in the number of moles of gas.

Step-by-Step Solution:

Step 1: The reaction is $S{O_2}(g) + \frac{1}{2}O_2(g) \rightleftharpoons S{O_3}(g)$ . This is an exothermic reaction, meaning it releases heat.

Step 2: According to Le Chatelier's principle, for an exothermic reaction, decreasing the temperature favors the forward reaction (formation of $S{O_3}$ ).

Step 3: The number of moles of gas decreases from 1.5 (1 mole $S{O_2}$ + 0.5 mole $O_2$ ) to 1 (1 mole $S{O_3}$ ). Increasing the pressure favors the side with fewer moles of gas.

Step 4: Therefore, low temperature and high pressure favor the formation of $S{O_3}$ .

Why other options are wrong:
- Option A: Increasing temperature would shift the equilibrium towards the reactants (reverse reaction) because the reaction is exothermic.
- Option C: Increasing both temperature and pressure would have opposing effects, with temperature disfavoring the product.
- Option D: Reducing both temperature and pressure would have opposing effects, with pressure disfavoring the product.

Key Concepts:
- Le Chatelier's principle
- Exothermic reactions
- Effect of pressure and temperature on equilibrium

Memory Aids:
- Exothermic: heat is a product, so decreasing temperature favors product formation.
- Pressure favors the side with fewer moles of gas.

Quick Tips:
- Apply Le Chatelier's principle to determine the effect of changes in conditions on equilibrium.

99 Easy

99 of 114Question permalink

Understanding +4 | -1

1 mol of

C{H_4}

contains

View answer and explanation

Correct option: B

4 g.atom of Hydrogen

Option B (4 g.atom of Hydrogen) is correct because 1 mole of

CH_4

contains 4 moles of hydrogen atoms, which is equivalent to 4 g.atom of Hydrogen.

Solution approach

Correct Answer + Core Reason: 4 g.atom of Hydrogen. 1 mol of $CH_4$ contains 4 moles of H atoms.

Step-by-Step Solution:

Step 1: Write the chemical formula for methane: $CH_4$ .

Step 2: One molecule of methane contains 1 carbon atom and 4 hydrogen atoms.

Step 3: Therefore, one mole of methane contains 1 mole of carbon atoms and 4 moles of hydrogen atoms.

Step 4: 4 moles of hydrogen atoms is equivalent to 4 g.atom of hydrogen.

Why other options are wrong:
- Option A: $6.02 \times {10^{23}}$ atoms of H is incorrect. It represents 1 mole of H atoms, not 4 moles.
- Option C: $1.81 \times {10^{23}}$ molecules of $C{H_4}$ is incorrect. 1 mole of $CH_4$ contains $6.022 \times 10^{23}$ molecules.
- Option D: 3.0 g of carbon is incorrect. 1 mole of carbon is 12g.

Key Concepts:
- Mole concept
- Chemical formula
- Gram atom

Memory Aids:
- Remember to multiply the number of atoms in the formula by the number of moles.

Quick Tips:
- Always pay attention to the chemical formula when calculating the number of atoms or moles.

100 Easy

100 of 114Question permalink

Application +4 | -1

Equivalent mass of

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}

in its reaction with

{\mathrm{I}_2}

is equal to :

View answer and explanation

Correct option: A

Molar\mathrm{ }mass

Option A (

Molar\mathrm{ }mass

) is correct because in the reaction of sodium thiosulfate with iodine, each mole of Na2S2O3 reacts with one mole of I2, so the equivalent mass is equal to the molar mass.

Solution approach

Correct Answer + Core Reason: $Molar\mathrm{ }mass$ . In the reaction with $I_2$ , each mole of $Na_2S_2O_3$ reacts with one mole of $I_2$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the reaction between $Na_2S_2O_3$ and $I_2$ :
$2Na_2S_2O_3 + I_2 \to Na_2S_4O_6 + 2NaI$ \nHowever, for equivalent mass calculation, we consider the change in oxidation state per mole of the reactant.
$Na_2S_2O_3 + \frac{1}{2}I_2 \to \frac{1}{2}Na_2S_4O_6 + NaI$

Step 2: Determine the change in oxidation state of sulfur in $Na_2S_2O_3$ .\nIn $Na_2S_2O_3$ , the average oxidation state of sulfur is +2.\nIn $Na_2S_4O_6$ , the average oxidation state of sulfur is +2.5.

Step 3: Calculate the n-factor (number of electrons transferred per mole of $Na_2S_2O_3$ ).\nThe change in oxidation state per sulfur atom is 0.5, and there are two sulfur atoms in $Na_2S_2O_3$ , so the n-factor is 1 (2 * 0.5 = 1).

Step 4: Calculate the equivalent mass using the formula: Equivalent mass = Molar mass / n-factor\nEquivalent mass = Molar mass / 1 = Molar mass

Step 5: Final Answer: $Molar\mathrm{ }mass$

Why other options are wrong:
- Option B: Incorrect, as the n-factor is 1.
- Option C: Incorrect, as the n-factor is 1.
- Option D: Incorrect, as the n-factor is 1.

Key Concepts:
- Equivalent mass
- Redox reactions
- Oxidation states
- n-factor

Memory Aids:
- Equivalent mass = Molar mass / n-factor

Quick Tips:
- Determine the n-factor by calculating the change in oxidation state per mole of the reactant.

101 Easy

101 of 114Question permalink

Remembering +4 | -1

The hardness of water is usually expressed in:

View answer and explanation

Correct option: A

ppm

Option A (ppm) is correct because the hardness of water is commonly expressed in parts per million (ppm).

Solution approach

Correct Answer + Core Reason: ppm because the hardness of water is commonly expressed in parts per million (ppm).

Step-by-Step Solution:

Step 1: Define hardness of water: Hardness of water is a measure of the amount of calcium and magnesium salts dissolved in water.

Step 2: Understand units of hardness: The hardness of water is usually expressed in terms of parts per million (ppm) of calcium carbonate ( $\mathrm{CaCO}_3$ ) equivalent.

Why other options are wrong:
- Option B: g/litre: While concentration can be expressed in g/litre, it's not the standard unit for water hardness.
- Option C: Mol/litre: Molarity is a measure of concentration, but not typically used to express water hardness.
- Option D: None of these: This is incorrect as ppm is a standard unit.

Key Concepts:
- Hardness of water
- Units of concentration
- Parts per million (ppm)

Memory Aids:
- Remember that ppm is the standard unit for water hardness.

Quick Tips:
- Know the common units used for expressing different properties of solutions.

102 Easy

102 of 114Question permalink

Remembering +4 | -1

One mole of a gas is defined as:

View answer and explanation

Correct option: D

The number of molecules in

22.4

litre of a gas at STP

Option D (The number of molecules in

22.4

litre of a gas at STP) is correct because one mole of any gas occupies 22.4 liters at Standard Temperature and Pressure (STP).

Solution approach

Correct Answer + Core Reason: The number of molecules in $22.4$ litre of a gas at STP. This is the definition of a mole of gas at STP.

Step-by-Step Solution:

Step 1: Recall the definition of a mole of gas at STP.\nOne mole of any gas occupies 22.4 liters at Standard Temperature and Pressure (STP).

Step 2: Relate the volume to the number of molecules.\nThis volume contains Avogadro's number of molecules ( $6.022 \times 10^{23}$ ).

Step 3: Conclusion.\nThe correct definition is the number of molecules in 22.4 liters of a gas at STP.

Why other options are wrong:
- Option A: The number of molecules in one liter of gas is not a fixed quantity and depends on temperature and pressure.
- Option B: The number of molecules in 2.24 liters of a gas is not equal to one mole.
- Option C: The number of atoms contained in 12g of C-14 isotope defines the mole based on carbon, not gas volume.

Key Concepts:
- Mole concept
- Standard Temperature and Pressure (STP)
- Avogadro's number

Memory Aids:
- Remember that 1 mole of any gas at STP occupies 22.4 L.

Quick Tips:
- Always remember the conditions for STP (0°C and 1 atm).

103 Easy

103 of 114Question permalink

Understanding +4 | -1

The mass of nitrogen per gram hydrogen in the compound hydrazine is exactly one and half times the mass of nitrogen in the compound ammonia. The fact illustrates the

View answer and explanation

Correct option: C

Law of multiple proportions

The correct answer is C (Law of multiple proportions) because the masses of nitrogen that combine with a fixed mass of hydrogen are in a simple whole-number ratio.

Solution approach

Correct Answer + Core Reason: Law of multiple proportions. The masses of nitrogen that combine with a fixed mass of hydrogen are in a simple whole-number ratio.

Step-by-Step Solution:

Step 1: Define the Law of Multiple Proportions: When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.

Step 2: In hydrazine, the mass of nitrogen per gram of hydrogen is 1.5 times that in ammonia.

Step 3: This implies that for a fixed mass of hydrogen, the ratio of nitrogen masses in the two compounds is 1.5:1 or 3:2, a simple whole number ratio.

Why other options are wrong:
- Option A: Law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. This is not directly illustrated by the given information.
- Option B: Multiple valency of nitrogen is a property of nitrogen but not the primary concept illustrated here.
- Option D: Law of definite proportions states that a given compound always contains exactly the same proportion of elements by mass. This doesn't explain the relationship between different compounds.

Key Concepts:
- Law of multiple proportions

Memory Aids:
- Multiple proportions: different compounds, simple ratio.

Quick Tips:
- Identify the elements and their mass ratios in different compounds.

104 Easy

104 of 114Question permalink

Remembering +4 | -1

A standard solution is one whose :

View answer and explanation

Correct option: C

Concentration is known

Option C is correct because a standard solution is defined as a solution whose concentration is accurately known.

Solution approach

Correct Answer + Core Reason: Concentration is known. A standard solution is one with a precisely determined concentration.

Step-by-Step Solution:

Step 1: Define a standard solution. A standard solution is a solution containing a precisely known concentration of an element or a substance.

Step 2: Consider the options:
A. Concentration is 1 M: This is a specific concentration, but a standard solution can have any known concentration.
B. Concentration is unknown: This is the opposite of what a standard solution is.
C. Concentration is known: This is the correct definition.
D. None of the above: Option C is correct.

Step 3: Conclusion: The correct answer is that the concentration is known.

Why other options are wrong:
- Option A: A standard solution is not necessarily 1 M. It just needs to have a known concentration.
- Option B: A standard solution must have a known concentration.
- Option D: Option C is correct, so this is incorrect.

Key Concepts:
- Standard solution
- Concentration

Memory Aids:
- Standard = Known

Quick Tips:
- Remember the definition of a standard solution.

105 Easy

105 of 114Question permalink

Remembering +4 | -1

The value of 1 amu is equal to

View answer and explanation

Correct option: D

Both (2) and (3)

Option D (Both (2) and (3)) is correct because 1 amu is equal to both

1.66 \times 10^{-27} \mathrm{~kg}

and

\frac{1}{N_A} \mathrm{~g}

, where

N_A

is Avogadro's number.

Solution approach

Correct Answer + Core Reason: Both (2) and (3). 1 amu is equal to $1.66 \times 10^{-27} \mathrm{~kg}$ and $\frac{{1}}{{N_A}} \mathrm{~g}$ .

Step-by-Step Solution:

Step 1: Recall the definition of 1 amu.
1 amu is defined as 1/12th the mass of a carbon-12 atom.

Step 2: State the value of 1 amu in kg.
1 amu = $1.66 \times 10^{-27} \mathrm{~kg}$ .

Step 3: Relate amu to Avogadro's number.
1 amu = $\frac{{1}}{{N_A}} \mathrm{~g}$ , where $N_A$ is Avogadro's number ( $6.022 \times 10^{23}$ ).

Step 4: Conclusion.\nTherefore, the value of 1 amu is equal to both $1.66 \times 10^{-27} \mathrm{~kg}$ and $\frac{{1}}{{N_A}} \mathrm{~g}$ .

Why other options are wrong:
- Option A: $1.66 \times 10^{-24} \mathrm{~kg}$ is incorrect; the exponent is wrong.
- Option B: $1.66 \times 10^{-27} \mathrm{~kg}$ is correct, but it's not the only correct option.
- Option C: $\frac{{1}}{{N_A}} \mathrm{~g}$ is correct, but it's not the only correct option.

Key Concepts:
- Atomic mass unit (amu)
- Avogadro's number

Memory Aids:
- Remember that 1 amu can be expressed in both kg and in terms of Avogadro's number.

Quick Tips:
- Be familiar with the relationship between amu, kg, and Avogadro's number.

106 Easy

106 of 114Question permalink

Analyzing +4 | -1

Assertion : Equivalent weight of

{Cu}

in

{CuO}

is 63.6 and in

{Cu}_2 {O} \,31.8

.

Reason : Equivalent weight of an element

=\frac{\text { Atomic weight of the element }}{\text { Valency of the element }}

View answer and explanation

Correct option: D

If assertion is false but reason is true.

Option D is correct. The assertion is false because the equivalent weight calculation for Cu in CuO is incorrect. The reason is true as it states the correct formula for equivalent weight.

Solution approach

Correct Answer + Core Reason: If assertion is false but reason is true.

Step-by-Step Solution:

Step 1: Analyze the assertion.\nThe assertion states the equivalent weight of Cu in CuO is 63.6 and in $Cu_2O$ is 31.8. We need to verify this.

Step 2: Analyze the reason.\nThe reason states the formula for equivalent weight: Equivalent weight = Atomic weight / Valency. This is a correct statement.

Step 3: Calculate the equivalent weight of Cu in CuO.\nIn CuO, the valency of Cu is 2. Atomic weight of Cu = 63.6 g/mol\nEquivalent weight of Cu in CuO = $\frac{{63.6}}{2} = 31.8$

Step 4: Calculate the equivalent weight of Cu in $Cu_2O$ .\nIn $Cu_2O$ , the valency of Cu is 1.\nEquivalent weight of Cu in $Cu_2O = \frac{{63.6}}{1} = 63.6$

Step 5: Compare the calculated values with the assertion.\nThe assertion states that the equivalent weight of Cu in CuO is 63.6, which is incorrect. The correct value is 31.8. The assertion states that the equivalent weight of Cu in $Cu_2O$ is 31.8, which is incorrect. The correct value is 63.6. Therefore, the assertion is false.

Step 6:

107 Easy

107 of 114Question permalink

Application +4 | -1

When

\mathrm{BrO}_3^ -

ion reacts with

\mathrm{B}{\mathrm{r}^ - }

ion in acidic solution

\mathrm{B}{\mathrm{r}_2}

is liberated. The equivalent weight of

\mathrm{KBr}{\mathrm{O}_3}

is :

View answer and explanation

Correct option: C

M/5

Option C (

M/5

) is correct because in the reaction of

BrO_3^-

with

Br^-

,

BrO_3^-

is converted to

Br_2

. The oxidation state of Br changes from +5 to 0, a change of 5. Therefore, the equivalent weight is M/5.

Solution approach

Correct Answer + Core Reason: $M/5$ . The change in oxidation state of Br is 5.

Step-by-Step Solution:

Step 1: Write the relevant half-reaction: $BrO_3^- \to Br_2$

Step 2: Determine the oxidation states of Br:
- In $BrO_3^-$ : Br has an oxidation state of +5.
- In $Br_2$ : Br has an oxidation state of 0.

Step 3: Calculate the change in oxidation state: The change is |+5 - 0| = 5.

Step 4: Calculate the equivalent weight: Equivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{Change in oxidation state}}}} = \frac{M}{5}$

Step 5: Final Answer: $M/5$

Why other options are wrong:
- Option A: This would be correct if the change in oxidation state was 8.
- Option B: This would be correct if the change in oxidation state was 3.
- Option D: This would be correct if the change in oxidation state was 6.

Key Concepts:
- Oxidation State
- Equivalent Weight

Memory Aids:
- Equivalent weight = (Molecular weight) / (Change in oxidation state)

Quick Tips:
- Always determine the oxidation states of the relevant elements.

108 Easy

108 of 114Question permalink

Remembering +4 | -1

The standard adopted for the determination of atomic weight of elements is based on

View answer and explanation

Correct option: B

{\mathrm{C}^{12}}

Option B (

{\mathrm{C}^{12}}

) is correct because the current standard for atomic weights is based on the carbon-12 isotope, which is assigned an atomic weight of exactly 12 amu.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}^{12}}$ . The standard for atomic weight is based on the carbon-12 isotope.

Step-by-Step Solution:

Step 1: Recall the standard for atomic weight.\nThe current standard for determining atomic weights is based on the carbon-12 isotope ( ${\mathrm{C}^{12}}$ ).

Step 2: Explain the standard.\nCarbon-12 is assigned an atomic weight of exactly 12 atomic mass units (amu).

Step 3: Conclusion.\nTherefore, the standard adopted for the determination of atomic weight of elements is based on ${\mathrm{C}^{12}}$ .

Why other options are wrong:
- Option A: ${\mathrm{H}^1}$ was an early standard, but it's no longer used.
- Option C: ${\mathrm{O}^{16}}$ was also a previous standard, but it has been replaced by carbon-12.
- Option D: ${\mathrm{S}^{32}}$ is not used as a standard for atomic weight determination.

Key Concepts:
- Atomic weight
- Atomic mass unit (amu)
- Carbon-12 isotope

Memory Aids:
- Remember that Carbon-12 is the current standard.

Quick Tips:
- Know the historical progression of atomic weight standards.

109 Easy

109 of 114Question permalink

Analyzing +4 | -1

In one experiment, 4 g of

H_2

combine with 32 g of

O_2

to form 36 g of

H_2O.

In another experiment, when 50 g of

H_2

combine with 400 g of

O_2

then 450 g of

H_2O

is formed. Above two experiments follow

View answer and explanation

Correct option: C

Both 1 & 2

Option C (Both 1 & 2) is correct because both experiments demonstrate the law of conservation of mass and the law of constant composition.

Solution approach

Correct Answer + Core Reason: Both 1 & 2. The experiments follow both the law of conservation of mass and the law of constant composition.

Step-by-Step Solution:

Step 1: Check the law of conservation of mass for both experiments:
- Experiment 1: 4 g $H_2$ + 32 g $O_2$ = 36 g $H_2O$ . Total mass of reactants = Total mass of products.
- Experiment 2: 50 g $H_2$ + 400 g $O_2$ = 450 g $H_2O$ . Total mass of reactants = Total mass of products.\nBoth experiments obey the law of conservation of mass.

Step 2: Check the law of constant composition for both experiments:
- Experiment 1: The mass ratio of H to O in $H_2O$ is 4/32 = 1/8.
- Experiment 2: The mass ratio of H to O in $H_2O$ is 50/400 = 1/8.\nThe mass ratio of H to O is constant in both experiments.

Step 3: Conclusion:\nBoth experiments follow the law of conservation of mass and the law of constant composition.

Why other options are wrong:
- Option A (The law of conservation of mass): This is incorrect because both laws are followed.
- Option B (The law of constant composition): This is incorrect because both laws are followed.
- Option D (Neither 1 nor 2): This is incorrect because both laws are followed.

Key Concepts:
- Law of Conservation of Mass
- Law of Constant Composition (Law of Definite Proportions)

Memory Aids:
- Law of Conservation of Mass: Mass is neither created nor destroyed in a chemical reaction.
- Law of Constant Composition: A given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.

Quick Tips:
- Check if the total mass of reactants equals the total mass of products to verify the law of conservation of mass.
- Check if the mass ratio of elements in a compound is constant to verify the law of constant composition.

110 Easy

110 of 114Question permalink

Understanding +4 | -1

Which of the following changes with increase in temperature?

View answer and explanation

Correct option: C

Fraction of solute present in water

The correct answer is C (Fraction of solute present in water) because temperature affects the volume of the solution, thus changing the concentration expressed as a fraction.

Solution approach

Correct Answer + Core Reason: Fraction of solute present in water. Temperature affects the volume of the solution.

Step-by-Step Solution:

Step 1: Consider the definitions of each concentration term.

Step 2: Molality is defined as moles of solute per kilogram of solvent. Mass is independent of temperature.

Step 3: Weight fraction of solute is the mass of solute divided by the total mass of the solution. Mass is independent of temperature.

Step 4: Mole fraction is the moles of solute divided by the total moles in the solution. Moles are independent of temperature.

Step 5: Fraction of solute present in water can be expressed as (volume of solute)/(volume of water). Volume is temperature-dependent.

Step 6: As temperature changes, the volume of water changes, thus affecting the fraction of solute present in water.

Why other options are wrong:
- Option A: Molality is defined as moles of solute per kg of solvent. Both moles and mass are independent of temperature.
- Option B: Weight fraction is the ratio of masses, which are independent of temperature.
- Option D: Mole fraction is the ratio of moles, which are independent of temperature.

Key Concepts:
- Concentration terms: molality, weight fraction, mole fraction
- Temperature dependence of volume

Memory Aids:
- Mass is conserved, volume is not.

Quick Tips:
- Remember that concentration units involving volume are temperature-dependent.

111 Easy

111 of 114Question permalink

Remembering +4 | -1

The formula which represents the simple ratio of atoms in a compound is called:

View answer and explanation

Correct option: C

Empirical formula

Option C (Empirical formula) is correct because the empirical formula represents the simplest whole-number ratio of atoms in a compound.

Solution approach

Correct Answer + Core Reason: Empirical formula. It represents the simplest whole number ratio of atoms in a compound.

Step-by-Step Solution:

Step 1: Define empirical formula.

Step 2: The empirical formula is the simplest positive integer ratio of atoms present in a compound.

Step 3: Contrast with molecular formula, which represents the actual number of atoms of each element in a molecule.

Why other options are wrong:
- Option A: Molecular formula represents the actual number of atoms of each element in a molecule, not the simplest ratio.
- Option B: Structure formula shows the arrangement of atoms in a molecule, not the ratio.
- Option D: Rational formula is an older term and not directly related to the simplest ratio of atoms.

Key Concepts:
- Empirical formula
- Molecular formula
- Chemical formulas

Memory Aids:
- Empirical = Simplest.

Quick Tips:
- To find the empirical formula, divide the subscripts in the molecular formula by their greatest common divisor.

112 Easy

112 of 114Question permalink

Analyzing +4 | -1

Few quantities with their units are given below. Mark the unit which is not correctly converted.

View answer and explanation

Correct option: C

1 \mathrm{~mg}=10^{10} \mathrm{~ng}

The correct answer is C (

1 \mathrm{~mg}=10^{10} \mathrm{~ng}

) because the conversion is incorrect. 1 mg is equal to 10⁶ ng, not 10¹⁰ ng.

Solution approach

Correct Answer + Core Reason: $1 \mathrm{~mg}=10^{10} \mathrm{~ng}$ is incorrect because the correct conversion is $1 \mathrm{~mg}=10^6 \mathrm{~ng}$ .

Step-by-Step Solution:

Step 1: Analyze option A: $1 \mathrm{~km}=10^6 \mathrm{~mm}$ .
$1 \mathrm{~km} = 1000 \mathrm{~m} = 1000 \times 100 \mathrm{~cm} = 1000 \times 100 \times 10 \mathrm{~mm} = 10^6 \mathrm{~mm}$ . This is correct.

Step 2: Analyze option B: $1 \mathrm{~mg}=10^{-6} \mathrm{~kg}$ .
$1 \mathrm{~mg} = 10^{-3} \mathrm{~g} = 10^{-3} \times 10^{-3} \mathrm{~kg} = 10^{-6} \mathrm{~kg}$ . This is correct.

Step 3: Analyze option C: $1 \mathrm{~mg}=10^{10} \mathrm{~ng}$ .
$1 \mathrm{~mg} = 10^{-3} \mathrm{~g} = 10^{-3} \times 10^9 \mathrm{~ng} = 10^6 \mathrm{~ng}$ . This is incorrect.

Step 4: Analyze option D: $1 \mathrm{~mL}=10^{-3} \mathrm{~dm}^3$ .
$1 \mathrm{~mL} = 1 \mathrm{~cm}^3 = (10^{-1} \mathrm{~dm})^3 = 10^{-3} \mathrm{~dm}^3$ . This is correct.

Why other options are wrong:
- Option A: $1 \mathrm{~km}=10^6 \mathrm{~mm}$ is a correct conversion.
- Option B: $1 \mathrm{~mg}=10^{-6} \mathrm{~kg}$ is a correct conversion.
- Option D: $1 \mathrm{~mL}=10^{-3} \mathrm{~dm}^3$ is a correct conversion.

Key Concepts:
- Unit conversions

Memory Aids:
- King Henry Died By Drinking Chocolate Milk (kilo, hecto, deca, base, deci, centi, milli)

Quick Tips:
- Always write out the conversion factors to avoid mistakes.

113 Easy

113 of 114Question permalink

Understanding +4 | -1

Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false.
(i) Gay-Lussac’s law of gaseous volumes is actually the law of definite proportion by volume.
(ii) Law of conservation of mass is true for physical change, but not for chemical change.
(iii) The percentage of oxygen in

H_2O_2

is different from that in

H_2O.

Hence, it violates law of definite proportions.
(iv) Fixed mass of A reacts with two different masses of B (say x and y), then the ratio of x/y can be any positive integer.
(v) At STP, 5 mL of

N_2

and

H_2

have different no. of molecules.

View answer and explanation

Correct option: C

TFFTF

Option C (TFFTF) is correct because the correct sequence of True/False for the given statements is TFFTF.

Solution approach

Correct Answer + Core Reason: TFFTF. The correct order of true/false statements is TFFTF.

Step-by-Step Solution:

Step 1: Analyze each statement individually.

Step 2: (i) Gay-Lussac’s law of gaseous volumes is actually the law of definite proportion by volume. This statement is TRUE.

Step 3: (ii) Law of conservation of mass is true for physical change, but not for chemical change. This statement is FALSE. Law of conservation of mass is true for both physical and chemical changes.

Step 4: (iii) The percentage of oxygen in $H_2O_2$ is different from that in $H_2O.$ Hence, it violates law of definite proportions. This statement is FALSE. Different compounds have different compositions, and this doesn't violate the law of definite proportions.

Step 5: (iv) Fixed mass of A reacts with two different masses of B (say x and y), then the ratio of x/y can be any positive integer. This statement is TRUE. This describes the law of multiple proportions.

Step 6: (v) At STP, 5 mL of $N_2$ and $H_2$ have different no. of molecules. This statement is FALSE. At STP, equal volumes of all gases contain equal numbers of molecules (Avogadro's Law).

Step 7: Combine the truth values: TFFTF.

Why other options are wrong:
- Option A: TTFTF is incorrect. The second and third statements are false.
- Option B: FTTFT is incorrect. The first statement is true, and the second and third are false.
- Option D: TFTTF is incorrect. The second and third statements are false, and the last statement is false.

Key Concepts:
- Gay-Lussac's Law
- Law of Conservation of Mass
- Law of Definite Proportions
- Law of Multiple Proportions
- Avogadro's Law

Memory Aids:
- Remember the fundamental laws of chemical combination.

Quick Tips:
- Carefully analyze each statement before assigning a truth value.

114 Easy

114 of 114Question permalink

Understanding +4 | -1

Assertion (A) : Molality is independent of temperature.

Reason (R) : Volume of solution does not change with temperature.

View answer and explanation

Correct option: C

\mathbf{A}

is true but

\mathbf{R}

is false.

Option C (

\mathbf{A}

is true but

\mathbf{R}

is false) is correct because molality is independent of temperature, but the reason given, that volume of solution does not change with temperature, is false.

Solution approach

Correct Answer + Core Reason: $\mathbf{A}$ is true but $\mathbf{R}$ is false. Molality is independent of temperature, but volume does change with temperature.

Step-by-Step Solution:

Step 1: Analyze the assertion (A). Molality is defined as moles of solute per kilogram of solvent. Since both moles and mass are independent of temperature, molality is also independent of temperature. Therefore, A is true.

Step 2: Analyze the reason (R). The volume of a solution changes with temperature due to thermal expansion. Therefore, R is false.

Step 3: Determine if R is the correct explanation of A. Since R is false, it cannot be the correct explanation of A.

Why other options are wrong:
- Option A: Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$ is incorrect because R is false.
- Option B: Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$ is incorrect because R is false.
- Option D: $\mathbf{A}$ is false but $\mathbf{R}$ is true is incorrect because A is true and R is false.

Key Concepts:
- Molality
- Temperature dependence of concentration
- Thermal expansion

Memory Aids:
- Remember that molality is based on mass, not volume.

Quick Tips:
- Distinguish between concentration units that depend on volume and those that depend on mass.

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