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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 301-330 of 436 questions

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301 Easy
301 of 436
Remembering +4 | -1
The net charge on ferrous ion is:
View answer and explanation
Correct option: A
+2+ 2
Option A (+2+ 2) is correct because the ferrous ion is Fe2+\mathrm{Fe}^{2+}, indicating a +2 charge.
302 Hard
302 of 436
Application +4 | -1
1 mol each of H3PO2,H3PO3{H}_3 {PO}_2, {H}_3 {PO}_3 and H3PO4{H}_3 {PO}_4 will neutralize x molx \mathrm{~mol} of NaOH,y mol{NaOH}, y \mathrm{~mol} of Ca(OH)2{Ca}({OH})_2 and z mol{z} \mathrm{~mol} of Al(OH)3.x,y{Al}({OH})_3 . x, y and zz are in the ratio
View answer and explanation
Correct option: D
1:1:11: 1: 1
Option B (4.78×10−11 mol L−14.78 \times 10^{-11} \mathrm{~mol} \mathrm{~L}^{-1}) is correct because it is the concentration of CO3-2 calculated using the given pH, K1, and K2 values for carbonic acid.
303 Easy
303 of 436
Remembering +4 | -1
Which represents per cent by weight?
View answer and explanation
Correct option: A
wt.  of  solutewt.  of  solution×100\frac{{\mathrm{wt}\mathrm{. }\;\mathrm{of}\;\mathrm{ solute}}}{{\mathrm{wt}\mathrm{. }\;\mathrm{of}\;\mathrm{ solution}}} \times 100
Option A is correct because percent by weight is defined as the weight of the solute divided by the weight of the solution, multiplied by 100.
304 Medium
304 of 436
Application +4 | -1
If 65.5 g of a metal chloride contains 35.5 g chlorine, then equivalent mass of metal is
View answer and explanation
Correct option: A
30
Option A (30) is correct because the equivalent mass of the metal is calculated using the ratio of masses and equivalent masses of the metal and chlorine.
305 Medium
305 of 436
Application +4 | -1
Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation, 2H2O2(aq)→2H2O(l)+O2(g)2{\mathrm{H}_2}{\mathrm{O}_2}\left( {aq} \right) \to 2{\mathrm{H}_2}\mathrm{O}\left( l \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) under conditions where one mole of gas occupies 24 dm3,100cm3\mathrm{d}{\mathrm{m}^3},100\mathrm{c}{\mathrm{m}^3} of XMXM solution of H2O2{\mathrm{H}_2}{\mathrm{O}_2} produces 3 dm3\mathrm{d}{\mathrm{m}^3} of O2{\mathrm{O}_2}. Thus, XX is :
View answer and explanation
Correct option: A
2.5
Option A (2.5) is correct because the molarity X is calculated based on the volume of oxygen produced from the decomposition of hydrogen peroxide.
306 Medium
306 of 436
Application +4 | -1
A transition metal MM forms a volatile chloride which has a vapour density of 94.8 . If it contains 74.75%74.75 \% of chlorine the formula of the metal chloride will be
View answer and explanation
Correct option: C
MCl4{MCl}_4
Option C (MCl4{MCl}_4) is correct because the valency of the metal is calculated to be 4, leading to the formula MCl4{MCl}_4.
307 Hard
307 of 436
Application +4 | -1
The density of 3M3 \mathrm{M} solution of sodium chloride is 1.252 g mL−11.252 \mathrm{~g} \mathrm{~mL}^{-1}. The molality of the solution will be:

(molar mass, NaCl=58.5 g mol−1\mathrm{NaCl}=58.5 \mathrm{~g} \mathrm{~mol}^{-1} )
View answer and explanation
Correct option: C
2.79 m2.79 \mathrm{~m}
Option C (2.79 m) is correct because it is calculated using the relationship between molarity, molality, and density.
308 Hard
308 of 436
Remembering +4 | -1
In the given chemical reaction, colours of the Fe2+  and  Fe3+\mathrm{F}{\mathrm{e}^{\mathrm{2 + }}}\,\,\mathrm{and}\,\,\mathrm{ F}{\mathrm{e}^{\mathrm{3 + }}} ions, are respectively
5Fe2++MnO4−+8H+→Mn2++4H2O+5Fe3+\mathrm{5}F{e^{\mathrm{2 + }}}\mathrm{ + }M\mathrm{n}O_4^ - + 8{H^ + } \to M{n^{2 + }} + 4{H_2}O + 5F{e^{3 + }}
View answer and explanation
Correct option: D
green, yellow
Option D (green, yellow) is correct because Fe2+\mathrm{F}{\mathrm{e}^{\mathrm{2 + }}} ions are typically green in solution, while Fe3+\mathrm{F}{\mathrm{e}^{\mathrm{3 + }}} ions are typically yellow.
309 Medium
309 of 436
Application +4 | -1
50 mL of an aqueous solution of glucose contains 6.02×10226.02 \times {10^{22}} molecules. The concentration of solution is:
View answer and explanation
Correct option: D
2.0 M
Option D (2.0 M) is correct because it is the concentration calculated using the number of molecules and volume of the solution.
310 Medium
310 of 436
Application +4 | -1
Mass of one molecule of glucose is (NA=6×1023)\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}\right)
View answer and explanation
Correct option: D
3×10−22 g3 \times 10^{-22} \mathrm{~g}
Option D (3×10−22 g3 \times 10^{-22} \mathrm{~g}) is correct because it is calculated by dividing the molar mass of glucose by Avogadro's number.
311 Medium
311 of 436
Analyzing +4 | -1
For the reaction :

H3PO3+Ba(OH)2→BaHPO3+2H2O\mathrm{H}_3 \mathrm{PO}_3+\mathrm{Ba}(\mathrm{OH})_2 \rightarrow \mathrm{BaHPO}_3+2 \mathrm{H}_2 \mathrm{O}

The correct statement(s) is/are

(a) The equivalent mass of H3PO3\mathrm{H}_3 \mathrm{PO}_3 is 41.

(b) BaHPO3\mathrm{BaHPO}_3 is the acid salt

(c) BaHPO3\mathrm{BaHPO}_3 is normal salt

(d) 1 mole of H3PO3\mathrm{H}_3 \mathrm{PO}_3 is completely neutralized by 1.5 moles of Ba(OH)2\mathrm{Ba}(\mathrm{OH})_2
View answer and explanation
Correct option: D
a and c are correct
Options a and c are correct because the equivalent mass of H3PO3H_3PO_3 is 41, and BaHPO3BaHPO_3 is a normal salt.
312 Hard
312 of 436
Application +4 | -1
Chlorofluorocarbons such as CCl3F(M=137.5){CCl}_3 {F}(\mathrm{M}=137.5) and CCl2F2(M=121){CCl}_2 {F}_2(\mathrm{M}=121) have been linked to ozone depletion in Antarctica. As of 2004, these gases were found in 275 and 605 parts per trillion (1012),\left(10^{12}\right), by volume. What are the concentrations of these gases under conditions typical of Antarctica stratosphere (200 K(200 \mathrm{~K} and 0.08 atm)?(R=0.08 l-atm/K-mol)0.08 \mathrm{~atm}) ?(R=0.08 \, l \text{-atm/K-mol})
View answer and explanation
Correct option: A
[CCl3F]=1.375×10−12 mol l−1,[CCl2F2]=3.025×10−12 mol l−1\left[{CCl}_3 {F}\right]=1.375 \times 10^{-12} \mathrm{~mol} \, {l}^{-1},\left[{CCl}_2 {F}_2\right]=3.025 \times 10^{-12} \mathrm{~mol} \, l^{-1}
Option A ([CCl3F]=1.375×10−12 mol,l−1,[CCl2F2]=3.025×10−12 mol,l−1\left[{CCl}_3 {F}\right]=1.375 \times 10^{-12} \mathrm{~mol} \\, {l}^{-1},\left[{CCl}_2 {F}_2\right]=3.025 \times 10^{-12} \mathrm{~mol} \\, l^{-1}) is correct because it correctly calculates the concentrations using the ideal gas law and the given parts per trillion values.
313 Medium
313 of 436
Application +4 | -1
100 mL100 \mathrm{~mL} of N/5NaOH\mathrm{N} / 5 \mathrm{NaOH} will neutralize
View answer and explanation
Correct option: C
1.2368 g1.2368 \mathrm{~g} of H3BO3\mathrm{H}_3 \mathrm{BO}_3
Option C (1.2368 g1.2368 \text{{ g}} of H3BO3\mathrm{H}_3 \mathrm{BO}_3) is correct because it is calculated by equating the equivalents of NaOH and H3BO3, considering the n-factor of H3BO3 is 1.
314 Medium
314 of 436
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
The correct answer is B (0.005 mole of KMnO4\mathrm{KMn}{\mathrm{O}_4} are left) because the milliequivalents of KMnO4\mathrm{KMn}{\mathrm{O}_4} exceed those of H2O2{\mathrm{H}_2}{\mathrm{O}_2}, leaving excess KMnO4\mathrm{KMn}{\mathrm{O}_4}.
315 Hard
315 of 436
Application +4 | -1
The number of gram atoms of oxygen present in 0.3 gram mole of (COOH)2⋅2H2O({COOH})_2 \cdot 2 {H}_2 {O} is
View answer and explanation
Correct option: B
1.8
Option B (1.8) is correct because it's the number of gram atoms of oxygen calculated from the given moles of the compound.
316 Hard
316 of 436
Application +4 | -1
The mineral haematite is Fe2O3\mathrm{Fe}_2 \mathrm{O}_3. Haematite ore contains unwanted material called gangue in addition of Fe2O3\mathrm{Fe}_2 \mathrm{O}_3. If 5.0 kg5.0 \mathrm{~kg} of ore contains 2.78 kg2.78 \mathrm{~kg} of Fe\mathrm{Fe}, what per cent of ore is Fe2O3\mathrm{Fe}_2 \mathrm{O}_3?
View answer and explanation
Correct option: A
79.4%79.4 \%
Option A (79.4 %) is correct because it accurately calculates the percentage of Fe2O3 in the ore based on the given mass of Fe.
317 Hard
317 of 436
Application +4 | -1
If the weight of metal oxide is xg{xg} containing yg{yg} of oxygen, the equivalent weight of metal will be
View answer and explanation
Correct option: D
E=8(x−y)yE=\frac{8(x-y)}{y}
Option D (E=8(x−y)yE=\frac{{8(x-y)}}{{y}}) is correct because it represents the equivalent weight of the metal calculated from the ratio of the weight of the metal to the weight of oxygen multiplied by 8.
318 Medium
318 of 436
Application +4 | -1
When 10 ml10 \mathrm{~ml} of acetic acid (density =0.8 g/ml=0.8 \mathrm{~g} / \mathrm{ml}) is mixed with 40 ml40 \mathrm{~ml} of water (density =1 g/ml=1 \mathrm{~g} / \mathrm{ml}) at a certain temperature, the final solution is found to have a density of 9698 g/ml.\frac{96}{98} \mathrm{~g} / \mathrm{ml}. The percent change in total volume on mixing is
View answer and explanation
Correct option: A
2%2 \%
Option A (2%2 \%) is correct because it is the percent change in volume calculated from the initial and final volumes after mixing.
319 Medium
319 of 436
Application +4 | -1
At NTP 2.24L of a gas weight 3.65 g3.65 \mathrm{~g}. The mole of the gas in 18.25 g18.25 \mathrm{~g} is :-
View answer and explanation
Correct option: B
0.5
Option B (0.5) is correct because it is the number of moles calculated using the given volume and mass of the gas at NTP.
320 Hard
320 of 436
Application +4 | -1
If mAm_A gram of a metal AA displaces mBm_B gram of another metal BB from its salt solution and if the equivalent weights are EAE_A and EBE_B respectively then equivalent weight of AA can be expressed as:
View answer and explanation
Correct option: A
EA=mAmB×EBE_A=\frac{m_A}{m_B} \times E_B
Option A (EA=mAmB×EBE_A=\frac{m_A}{m_B} \times E_B) is correct because it directly applies Faraday's laws of electrolysis, which state that the mass of a substance deposited or dissolved is proportional to its equivalent weight.
321 Hard
321 of 436
Application +4 | -1
40ml40 \mathrm{ml} of 0.8 N0.8 \mathrm{~N} oxalic acid dihydrate is shaken with 0.75 g0.75 \mathrm{~g} active charcoal. The final concentration of the solution after adsorption is 0.2 N0.2 \mathrm{~N}. What is the amount of oxalic acid adsorbed per gm of carbon?
View answer and explanation
Correct option: D
2.02 g2.02 \mathrm{~g}
Option D (2.02 g2.02 \mathrm{~g}) is correct because it is the amount of oxalic acid adsorbed per gram of carbon, calculated from the change in concentration.
322 Medium
322 of 436
Analyzing +4 | -1
Assertion (A) : In the following half reaction:
Cr2O72−⟶Cr3+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow \mathrm{Cr}^{3+}
Equivalent weight of Cr2O72−= Molar mass 3\mathrm{Cr}_2 \mathrm{O}_7^{2-}=\frac{\text { Molar mass }}{3}
Reason (R) : Change in oxidation number of chromium is 3 per atom.
View answer and explanation
Correct option: D
A\mathbf{A} is false but R\mathbf{R} is true.
Option D is correct because the assertion is false; the equivalent weight is molar mass divided by 6, not 3. The reason is true; the change in oxidation number of chromium is 3 per atom.
323 Medium
323 of 436
Application +4 | -1
What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene
View answer and explanation
Correct option: C
9.6 kg
Option C (9.6 kg) is correct because it is the calculated mass of oxygen required for the complete combustion of 2.8 kg of ethylene based on stoichiometry.
324 Hard
324 of 436
Application +4 | -1
What is the mass of 3.01×10223.01 \times 10^{22} molecules of ammonia?
View answer and explanation
Correct option: A
0.85 g
Option A (0.85 g) is correct because it is calculated using the mole concept and Avogadro's number to find the mass of the given number of ammonia molecules.
325 Easy
325 of 436
Analyzing +4 | -1
An amount of 5 moles of A,\text{A,} 6 moles of B\text{B} and excess amount of C\mathrm{C} are mixed to produce a final product D\mathrm{D}, according to the following reactions.

A+2B→II+C→B+D\begin{aligned} & \mathrm{A}+2 \mathrm{B} \rightarrow \mathrm{I} \\ & \mathrm{I}+\mathrm{C} \rightarrow \mathrm{B}+\mathrm{D} \end{aligned}

What is the maximum moles of D,\mathrm{D}, which can be produced assuming that the products formed can also be reused in the reactions?
View answer and explanation
Correct option: C
5 moles
Option C (A is true but R is false) is correct because HBF4 is a strong acid, but the given reaction of B(OH)3 with HF doesn't explain why HBF4 is a strong acid.
326 Medium
326 of 436
Analyzing +4 | -1
In the following reaction, As2 S3+H++NO3−⟶NO+H2O+AsO43−+SO42−\mathrm{As}_2 \mathrm{~S}_3+\mathrm{H}^{+}+\mathrm{NO}_3^{-} \longrightarrow \mathrm{NO}+\mathrm{H}_2 \mathrm{O}+\mathrm{AsO}_4^{3-}+\mathrm{SO}_4^{2-} the equivalent weight of As2 S3\mathrm{As}_2 \mathrm{~S}_3 is related to its molecular weight by :
View answer and explanation
Correct option: D
M/28M / 28
Option D is correct because the n-factor is calculated based on the change in oxidation states of As and S, leading to M/28.
327 Medium
327 of 436
Analyzing +4 | -1
Assertion (A) : 1 mole of FeC2O4\mathrm{FeC}_2 \mathrm{O}_4 is oxidised by 0.6 mole of MnO4−\mathrm{MnO}_4^{-}in acidic medium.

Reason (R) : MnO4−\mathrm{MnO}_4^{-}oxidises both Fe2+\mathrm{Fe}^{2+} as well as C2O42−\mathrm{C}_2 \mathrm{O}_4^{2-}.
View answer and explanation
Correct option: A
Both A\mathbf{A} and R\mathbf{R} are true and R\mathbf{R} is the correct explanation of A\mathbf{A}.
Option A is correct because both the assertion and the reason are true, and the reason correctly explains why 1 mole of FeC2O4\mathrm{FeC}_2 \mathrm{O}_4 requires 0.6 moles of MnO4−\mathrm{MnO}_4^{-} for oxidation in acidic medium.
328 Hard
328 of 436
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 determined by calculating the moles of each component and finding the simplest whole number ratio, which corresponds to CuSO₄⋅(NH₄)₂SO₄⋅6H₂O.
329 Hard
329 of 436
Application +4 | -1
Mole fraction of solute in an aqueous solution of NaOH\mathrm{NaOH} is 0.1.0.1. If the specific gravity of the solution is 1.4,1.4, then the molarity of the solution is
View answer and explanation
Correct option: A
6.93
Option A (6.93) is correct because the molarity is calculated using the mole fraction, specific gravity, and molar masses of the solute and solvent.
330 Medium
330 of 436
Application +4 | -1
The element Q\mathrm{Q} has the following isotopic composition, Q200:80%;Q190:15%\mathrm{Q}^{200}: 80 \%; \mathrm{Q}^{190}: 15 \% and Q210:5%\mathrm{Q}^{210}: 5 \%. The average atomic mass of the naturally occurring element Q\mathrm{Q} is
View answer and explanation
Correct option: C
199 u199 \mathrm{~u}
Option C (199 u) is correct because it's calculated as the weighted average of the isotopic masses based on their abundances.
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