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NEET • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

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271 Easy
271 of 313
Remembering +4 | -1
One mole of a gas is defined as:
View answer and explanation
Correct option: D
The number of molecules in 22.422.4 litre of a gas at STP
Option D (The number of molecules in 22.422.4 litre of a gas at STP) is correct because one mole of any gas occupies 22.4 liters at Standard Temperature and Pressure (STP).
272 Easy
272 of 313
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.
273 Hard
273 of 313
Analyzing +4 | -1
1.020 g1.020 {~g} of metallic oxide contains 0.540 g0.540 {~g} of the metal. If the specific heat of the metal, MM is 0.216 cal deg−1 g−10.216\, {cal}\, {deg}^{-1} {~g}^{-1}, the molecular formula of its oxide is
View answer and explanation
Correct option: B
M2O3{M}_2 {O}_3
Option B (M2O3) is correct because the equivalent weight of the metal is calculated from its specific heat, and this is used to determine the metal's valency and hence the oxide's formula.
274 Hard
274 of 313
Understanding +4 | -1
Assertion: Equivalence point is a theoretical value.
Reason: End point is an experimental value.
View answer and explanation
Correct option: D
Both Assertion and Reason are correct but Reason is not correct explanation of Assertion
Option D (Both Assertion and Reason are correct but Reason is not correct explanation of Assertion) is correct because both statements are true, but the experimental nature of the endpoint doesn't explain why the equivalence point is theoretical.
275 Medium
275 of 313
Analyzing +4 | -1
(A): Empirical formula of glucose and acetic acid is CH2O\mathrm{CH}_2 \mathrm{O}.
(R): If percentage composition by mass of individual elements in different compounds are same, then their empirical formulae are same.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because the empirical formula is determined by the simplest whole number ratio of elements, and if the percentage composition is the same, the ratio will be the same.
276 Hard
276 of 313
Application +4 | -1
Which of the following is the correct empirical and molecular formulae of a compound, if the molecular mass of compound is 80 and compound contains 60% of C, 5% of H and 35% of N ?
View answer and explanation
Correct option: A
C2H2N ;  C4H4N2{C_2}{H_2}N\,;\,\,{C_4}{H_4}{N_2}
Option A (C2H2N    C4H4N2{C_2}{H_2}N\,\;\,{C_4}{H_4}{N_2}) is correct. The empirical formula is determined from the percentage composition, and the molecular formula is a multiple of the empirical formula based on the molecular mass.
277 Easy
277 of 313
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.
278 Easy
278 of 313
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 1.66×10−27 kg1.66 \times 10^{-27} \mathrm{~kg} and 1NA g\frac{1}{N_A} \mathrm{~g}, where NAN_A is Avogadro's number.
279 Hard
279 of 313
Analyzing +4 | -1
Cupric ammonium sulphate was found to contain
27.03%27.03\% water of crystallization by mass. Upon
strongly heating, it gave cupric oxide corresponding
to 19.89%19.89\% of starting mass. Find the empirical
formula of cupric ammonium sulphate. (Cu=63.5)(\text{Cu} = 63.5)
View answer and explanation
Correct option: A
CuSO4⋅(NH4)2SO4⋅6H2O\mathrm{CuSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}
Option A is correct because the empirical formula is CuSO4.(NH4)2SO4.6H2O, which is derived from the mass percentages of water and cupric oxide.
280 Medium
280 of 313
Application +4 | -1
The number of moles of K2Cr2O7{K_2}C{r_2}{O_7} that will be needed to react with one mole of nitrite ion in acidic solution is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because it is the correct stoichiometric ratio for the reaction between dichromate and nitrite ions in acidic solution.
281 Medium
281 of 313
Application +4 | -1
0.3 g of an oxalate salt was dissolved in 100 mL solution. The solution required 90 mL of N/20N/20 KMnO4\mathrm{KMn}{\mathrm{O}_4} for complete oxidation. The % of oxalate ion in salt is:
View answer and explanation
Correct option: B
66%66\%
Option B (66%66\%) is correct because the percentage of oxalate ion is calculated based on the stoichiometry of the reaction with potassium permanganate.
282 Easy
282 of 313
Analyzing +4 | -1
Assertion : Equivalent weight of Cu{Cu} in CuO{CuO} is 63.6 and in Cu2O 31.8{Cu}_2 {O} \,31.8.

Reason : Equivalent weight of an element = Atomic weight of the element  Valency of the 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.
283 Medium
283 of 313
Understanding +4 | -1
By Victor meyer’s method, one determine the vapour density if:
View answer and explanation
Correct option: C
Volatile liquid
Option C is correct because Victor Meyer's method is specifically used to determine the vapor density of volatile liquids.
284 Easy
284 of 313
Application +4 | -1
When BrO3−\mathrm{BrO}_3^ - ion reacts with Br−\mathrm{B}{\mathrm{r}^ - } ion in acidic solution Br2\mathrm{B}{\mathrm{r}_2} is liberated. The equivalent weight of KBrO3\mathrm{KBr}{\mathrm{O}_3} is :
View answer and explanation
Correct option: C
M/5M/5
Option C (M/5M/5) is correct because in the reaction of BrO3−BrO_3^- with Br−Br^-, BrO3−BrO_3^- is converted to Br2Br_2. The oxidation state of Br changes from +5 to 0, a change of 5. Therefore, the equivalent weight is M/5.
285 Easy
285 of 313
Remembering +4 | -1
The standard adopted for the determination of atomic weight of elements is based on
View answer and explanation
Correct option: B
C12{\mathrm{C}^{12}}
Option B (C12{\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.
286 Hard
286 of 313
Analyzing +4 | -1
Assertion:- Equivalent wt. of Cu\mathrm{Cu} in both CuO\mathrm{CuO} and Cu2O\mathrm{Cu}_2 \mathrm{O} is 31.25.
Reason:- Equivalent wt. of an element is constant.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
Option D (If both Assertion & Reason are False) is correct because the equivalent weight of Cu is different in CuO and Cu2O, and the equivalent weight of an element is not always constant.
287 Hard
287 of 313
Application +4 | -1
A 0.50 MM solution of KIKI reacts with excess of H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} and KIO3\mathrm{KI}{\mathrm{O}_3} solutions according to the equation, 6H++5I−+IO3−→3I2+3H2O6{\mathrm{H}^ + } + 5{\mathrm{I}^ - } + \mathrm{IO}_3^ - \to 3{\mathrm{I}_2} + 3{\mathrm{H}_2}\mathrm{O}. Which of the following statements is true?
View answer and explanation
Correct option: C
0.5 litre of the KIKI solution produces 0.15 mole of I2{\mathrm{I}_2}
Option D (0.5 litre of the KIKI solution produces 0.15 mole of I2{\mathrm{I}_2}) is correct because it is derived from the stoichiometry of the given reaction.
288 Easy
288 of 313
Analyzing +4 | -1
In one experiment, 4 g of H2H_2 combine with 32 g of O2O_2 to form 36 g of H2O.H_2O. In another experiment, when 50 g of H2H_2 combine with 400 g of O2O_2 then 450 g of H2OH_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.
289 Hard
289 of 313
Analyzing +4 | -1
A mixture of NaHC2O4\mathrm{NaHC}_2 \mathrm{O}_4 and KHC2O4⋅H2C2O4\mathrm{KHC}_2 \mathrm{O}_4 \cdot \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 required equal volumes of 0.2 N KMnO40.2 \mathrm{~N} \mathrm{~KMnO}_4 and 0.12 NNaOH0.12 \mathrm{~N} \mathrm{NaOH} separately. What is the molar ratio of NaHC2O4\mathrm{NaHC}_2 \mathrm{O}_4 and KHC2O4⋅H2C2O4\mathrm{KHC}_2 \mathrm{O}_4 \cdot \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 in the mixture?
View answer and explanation
Correct option: D
3:13: 1
Option D is correct because the molar ratio of NaHC2O4 to KHC2O4.H2C2O4 is 3:1, calculated by equating the equivalents of KMnO4 and NaOH used in the titrations.
290 Hard
290 of 313
Application +4 | -1
The ratio of amounts of H2S{\mathrm{H}_2}\mathrm{S} needed to precipitate all the metal ions from 100 mL of 1 M AgNO3\mathrm{AgN}{\mathrm{O}_3} and 100 mL of CuSO4,\mathrm{CuS}{\mathrm{O}_4}, will be
View answer and explanation
Correct option: B
1:21\mathrm{ }:\mathrm{ }2
Option B (1:21\mathrm{ }:\mathrm{ }2) is correct because AgNO3AgNO_3 requires 1 mole of H2SH_2S per mole of AgNO3AgNO_3 while CuSO4CuSO_4 requires 1 mole of H2SH_2S per mole of CuSO4CuSO_4, and since the molarity and volume are the same, the ratio is 1:2.
291 Hard
291 of 313
Analyzing +4 | -1
3L mixture of O2(g){O}_2(\mathrm{g}) and N2(g){N}_2(\mathrm{g}) is passed through ozonizer. The gas mixture obtained contains O2,O3{O}_2, {O}_3 and N2{N}_2 gases only. This obtained mixture is passed through alkaline pyrogallol and then through turpentine oil, there were 1100 mL1100 \mathrm{~mL} and 600 mL600 \mathrm{~mL} volume contractions respectively. The volume of O2(g){O}_2(\mathrm{g}) and N2(g){N}_2(\mathrm{g}) in the original mixture respectively is
View answer and explanation
Correct option: D
2 L,1 L2 \mathrm{~L}, 1 \mathrm{~L}
Option D (2 L,1 L2 \mathrm{~L}, 1 \mathrm{~L}) is correct because the volume contraction with alkaline pyrogallol represents the volume of O2O_2 and the subsequent contraction with turpentine oil represents the volume of O3O_3 formed from O2O_2.
292 Easy
292 of 313
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.
293 Medium
293 of 313
Understanding +4 | -1
The weight of a substance that displaces 22.4 litre air at NTP is:
View answer and explanation
Correct option: A
Mol. wt.
Option A is correct because the weight of a substance that displaces 22.4 liters of air at NTP is equal to its molecular weight.
294 Medium
294 of 313
Analyzing +4 | -1
25 mL of 0.50 MH2O2M{\mathrm{H}_2}{\mathrm{O}_2} solution is added to 50 mL of 0.20 M MKMnO4M\mathrm{KMn}{\mathrm{O}_4} in acidic solution. Which of the following statements is true?
View answer and explanation
Correct option: B
0.005 mole of KMnO4\mathrm{KMn}{\mathrm{O}_4} are left
Option B (0.005 mole of KMnO4 are left) is correct because KMnO4 is in excess after the reaction with H2O2.
295 Medium
295 of 313
Understanding +4 | -1
The number of water molecules is maximum in:
View answer and explanation
Correct option: B
18 moles of water
Option B (18 moles of water) is correct because the number of water molecules is directly proportional to the number of moles, and 18 moles is the largest quantity.
296 Easy
296 of 313
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.
297 Easy
297 of 313
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 mg=1010 ng1 \mathrm{~mg}=10^{10} \mathrm{~ng}
The correct answer is C (1 mg=1010 ng1 \mathrm{~mg}=10^{10} \mathrm{~ng}) because the conversion is incorrect. 1 mg is equal to 106 ng, not 1010 ng.
298 Medium
298 of 313
Remembering +4 | -1
Dipole-dipole interaction energy between rotating polar molecules is proportional to (rr is distance between polar molecules)
View answer and explanation
Correct option: D
1r6\frac{1}{r^6}
Option D (1r6\frac{1}{r^6}) is correct because the dipole-dipole interaction energy between rotating polar molecules is proportional to 1r6\frac{1}{r^6}, where r is the distance between the molecules.
299 Hard
299 of 313
Remembering +4 | -1
A chemical balance used normally for weighing in laboratory can weigh upto a least count of :
View answer and explanation
Correct option: C
0.0002 g
Option C is correct because a typical chemical balance in a laboratory can measure up to a least count of 0.0002 g.
300 Medium
300 of 313
Analyzing +4 | -1
Assertion: For calculating the molality or the mole fraction of solute, if the molarity is known, it is necessary to know the density of the solution.

Reason: Molality, molarity and the mole fraction of solute can be calculated from the weight percentage and the density of the solution.
View answer and explanation
Correct option: B
Assertion is true, Reason is true and Reason is not correct explanation for Assertion
Option B is correct because both the assertion and reason are true, but the reason doesn't directly explain the assertion.
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