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JEE • CHEMISTRY • SOLUTIONS

Solutions 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 241-270 of 460 questions

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241 Easy
241 of 460
Understanding +4 | -1
In cold countries, ethylene glycol is added to water in the radiators of cars during winters. It results in :
View answer and explanation
Correct option: D
Lowering in freezing point
Option D (Lowering in freezing point) is correct because ethylene glycol acts as an antifreeze, lowering the freezing point of water.
242 Medium
242 of 460
Understanding +4 | -1
Which of the following statements is true about Henry's law?
View answer and explanation
Correct option: B
The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution
Option B (The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution) is correct because it accurately describes Henry's Law.
243 Medium
243 of 460
Understanding +4 | -1
The pair of boiling point and compound are given as,

C6H6CH3OHC6H5NO2C6H5NH280∘C65∘C212∘C184∘C(I)(II)(III)(IV)\begin{array}{cccccccccccccccccccc} {{C_6}{H_6}}&{C{H_3}OH}&{{C_6}{H_5}N{O_2}}&{{C_6}{H_5}N{H_2}} \\ {80^\circ C}&{65^\circ C}&{212^\circ C}&{184^\circ C} \\ {\left( {\text{I}} \right)}&{\left( {{\text{II}}} \right)}&{\left( {{\text{III}}} \right)}&{\left( {{\text{IV}}} \right)} \end{array}

Which will show lowest vapour pressure at room temperature?
View answer and explanation
Correct option: C
C6H5NO2{{C_6}{H_5}N{O_2}}
Option C (C6H5NO2{{C_6}{H_5}N{O_2}}) is correct because the compound with the highest boiling point will have the lowest vapor pressure at a given temperature.
244 Medium
244 of 460
Application +4 | -1
35.4 mL of HCl is required for the neutralisation of a solution containing 0.275 g of sodium hydroxide. The normality of hydrochloric acid is
View answer and explanation
Correct option: C
0.194  N0.194\;N
Option C (0.194 N) is correct because it is the normality of the hydrochloric acid calculated using the neutralization reaction with sodium hydroxide.
245 Easy
245 of 460
Application +4 | -1
3.01×10233.01 \times 10^{23} molecules of glucose are present in 500 mL500 \mathrm{~mL} of solution. The molarity of the solution is
View answer and explanation
Correct option: B
1 M
Option B (1 M) is correct because the molarity is calculated by dividing the number of moles of glucose by the volume of the solution in liters.
246 Medium
246 of 460
Application +4 | -1
Addition of 0.643 g0.643 \mathrm{~g} of a compound to 50 mL50 \mathrm{~mL} of benzene (density =0.879 g/mL)=0.879 \mathrm{~g} / \mathrm{mL}) lowers the freezing point from 5.51∘C5.51^{\circ} \mathrm{C} to 5.03∘C5.03^{\circ} \mathrm{C}. If Kf\mathrm{K}_{\mathrm{f}} for benzene is 5.12 K kg mol–1,5.12 \mathrm{~K ~kg ~mol}^{–1}, calculate the molecular weight of the compound.
View answer and explanation
Correct option: A
156
Option A (156) is correct because the calculated molecular weight of the compound using the freezing point depression formula is approximately 156.
247 Easy
247 of 460
Application +4 | -1
One mole of non-volatile solute is dissolved in two mole of water. The vapour pressure of the solution relative to that of water is :
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Correct option: A
2/32/3
Option A (2/32/3) is correct because it is calculated using Raoult's Law, which relates the vapor pressure of the solution to the mole fraction of the solvent.
248 Medium
248 of 460
Understanding +4 | -1
Assertion (A): Aquatic species are more comfortable in cold water rather than in warm water.

Reason (R): The solubility of gas O2{O}_2, increases with decrease of temperature due to increase of KH.{K}_{{H}}.
View answer and explanation
Correct option: A
Both A & R are true, R is the correct explanation of A
Option A is correct because aquatic species are more comfortable in cold water due to the increased solubility of oxygen at lower temperatures, which is related to Henry's Law.
249 Medium
249 of 460
Remembering +4 | -1
The van't Hoff factor (i) for a dilute aqueous solution of glucose is:
View answer and explanation
Correct option: B
1
Option B (1) is correct because glucose is a non-electrolyte and does not dissociate into ions in solution; therefore, its van't Hoff factor is 1.
250 Easy
250 of 460
Understanding +4 | -1
Negative deviation from Raoult's law is law is observed in which one of the following binary liquid mixtures?
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Correct option: C
Acetone and chloroform
Option C (Acetone and chloroform) is correct because they form hydrogen bonds, leading to stronger intermolecular forces and a decrease in vapor pressure.
251 Medium
251 of 460
Application +4 | -1
Phenol dimerizes in benzene. If the observed molecular mass of phenol in solution is 120, then its degree of dimerization is
View answer and explanation
Correct option: B
0.433
Option B (0.433) is correct because it's calculated using the observed molecular mass and the formula for the degree of dimerization.
252 Hard
252 of 460
Application +4 | -1
0.010M0.010 \mathrm{M} solution an acid HAH A freezes at −0.0205∘C-0.0205^{\circ} \mathrm{C}. If Kf\mathrm{K}_{\mathrm{f}} for water is 1.860 K kg mol−11.860 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, the ionization constant of the conjugate base of the acid will be (assume 0.010M=0.010 m0.010 \mathrm{M}=0.010 \mathrm{~m} )
View answer and explanation
Correct option: C
9.0×10−119.0 \times 10^{-11}
Option C (9.0×10−119.0 \times 10^{-11}) is correct because it involves calculating the van't Hoff factor, the degree of ionization, the acid dissociation constant, and finally, the base dissociation constant using the relationship between KaK_a, KbK_b, and KwK_w.
253 Medium
253 of 460
Understanding +4 | -1
Which of the following acid has the same molecular weight and equivalent weight.
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Correct option: A
H3PO2{H_3}P{O_2}
Option A (H3PO2{H_3}P{O_2}) is correct because it is a monoprotic acid, meaning its n-factor is 1, resulting in the same molecular and equivalent weight.
254 Hard
254 of 460
Application +4 | -1
A solution is made by mixing 300 ml. of 1.5 M Al2(SO4)3A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}} + 300 ml. of 2 M CaSO4CaS{{O}_{4}} + 400 ml. of 3.5 M CaCl2.CaC{{l}_{2}}. Find final molarity of

(a) SO4−2{SO}_4^{-2}\quad (b) Ca2+{Ca}^{2+}

[Assume complete dissociation of these compounds]
View answer and explanation
Correct option: A
(a) 1.95 M;1.95 \mathrm{~M}; (b) 2 M2 \mathrm{~M}
Option A ((a) 1.95 M;1.95 \mathrm{~M}; (b) 2 M2 \mathrm{~M}) is correct because it correctly calculates the final molarities of sulfate and calcium ions considering the total moles and total volume of the solution.
255 Easy
255 of 460
Remembering +4 | -1
The solubility of a gas in water depends on :
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the solubility of a gas in water is affected by the nature of the gas, temperature, and pressure of the gas.
256 Medium
256 of 460
Application +4 | -1
What is the amount of CaCl2(i=2.9)\mathrm{CaCl}_2(\mathrm{i}=2.9) dissolved in 4 litre of water such that its osmotic pressure is 1 atm at 27∘C27^{\circ} \mathrm{C}?
View answer and explanation
Correct option: B
6.21 g
Option B (6.21 g) is correct because it is calculated using the modified Van't Hoff equation, accounting for the van't Hoff factor (i) and solving for the number of moles of CaCl2\mathrm{CaCl}_2 dissolved.
257 Easy
257 of 460
Understanding +4 | -1
An unriped mango placed in a concentrated salt solution to prepare pickle, shrivels because …….. .
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Correct option: D
It loses water due to onions
The correct answer is that the mango loses water due to osmosis. Osmosis is the movement of water from a region of high water concentration (inside the mango) to a region of low water concentration (the concentrated salt solution).
258 Medium
258 of 460
Understanding +4 | -1
Which of the following statements is incorrect?
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Correct option: D
The minimum amount of solute dissolved in a given amount of solvent is its solubility.
Option D is correct because solubility refers to the maximum, not minimum, amount of solute that can dissolve in a given amount of solvent.
259 Medium
259 of 460
Application +4 | -1
Which of the following will have the highest boiling point at 1 atm1 {~atm} pressure
View answer and explanation
Correct option: C
0.1 M BaCl20.1 \,{M}\, {BaCl}_2
The correct answer is 0.1 M BaCl2 because it has the highest van't Hoff factor (i=3) among the given options, leading to the greatest elevation in boiling point.
260 Medium
260 of 460
Application +4 | -1
Calculate the mole fractions of the components of the solution composed by 92 g92 \mathrm{~g} glycerol and 90 g90 \mathrm{~g} water?  (M? \, \, (M (water)=18;M()=18 ; M(glycerol)=92))=92)
View answer and explanation
Correct option: A
0.833;0.1670.833; 0.167
Option A (0.833; 0.167) is correct because it accurately calculates the mole fractions of water and glycerol based on their respective masses and molar masses.
261 Medium
261 of 460
Analyzing +4 | -1
Statement I: Molarity of 0.02 N HNO30.02 \mathrm{~N} \mathrm{~HNO}_3 is 0.02 M0.02 \mathrm{~M}.

Statement II: Molarity and normality of a solution are always same.
View answer and explanation
Correct option: C
If Statement I is correct but Statement II is false
Option C (If Statement I is correct but Statement II is false) is correct because Statement I is true for HNO3HNO_3 (n-factor=1), but molarity and normality are not always the same; they are related by the n-factor.
262 Medium
262 of 460
Application +4 | -1
Percent by mass of solute (molar mass =25 g/mol=25 \mathrm{~g} / \mathrm{mol} ) in its aqueous solution is 30 . The mole fraction and molality of the solute in solution are
View answer and explanation
Correct option: B
Mole fraction =0.236=0.236, Molality =17.14=17.14
Option B (Mole fraction =0.236, Molality =17.14) is correct because it is the mole fraction and molality calculated from the given mass percentage of the solute.
263 Medium
263 of 460
Application +4 | -1
The vapour pressure of a pure liquid A is 40 mm Hg at 310 K. The vapour pressure of this liquid in a solution with liquid B is 32 mm Hg. What is the mole fraction of A in the solution if it obeys the Raoult’s law?
View answer and explanation
Correct option: D
0.8
Option D (0.8) is correct because Raoult's law states that the partial pressure of a component in a solution is the product of its mole fraction and its vapor pressure in the pure state.
264 Hard
264 of 460
Application +4 | -1
Molal depression constant for a solvent is 4.0K  kg  mol−14.0\mathrm{ }K\mathrm{ }\,\,kg\mathrm{ }\,\,mo{l^{ - 1}}. The depression in the freezing point of the solvent for 0.03  mol  kg−10.03\mathrm{ }\,\,mol\,\,\mathrm{ k}{\mathrm{g}^{ - 1}} solution of K2SO4{K_2}S{O_4} is (Assume complete dissociation of the electrolyte)
View answer and explanation
Correct option: B
0.36 K
Option B (0.36 K) is correct because the depression in freezing point is calculated using the formula ΔTf = iKfm, considering the van't Hoff factor for K2SO4.
265 Easy
265 of 460
Remembering +4 | -1
For dilute solution Raoult’s law states that
View answer and explanation
Correct option: A
The relative lowering of vapour pressure is equal to mole fraction of solute
Option A (The relative lowering of vapour pressure is equal to mole fraction of solute) is correct because that is the definition of Raoult's law for dilute solutions.
266 Medium
266 of 460
Application +4 | -1
When 45 gm solute is dissolved in 600 gm water freezing point lower by 2.2 K, calculate molar mass of solute (Kf=1.86 K kg)\left( {{K_f} = 1.86\,K\,kg} \right)
View answer and explanation
Correct option: A
63.4
Option A (63.4) is correct because the molar mass of the solute is calculated using the freezing point depression formula.
267 Medium
267 of 460
Application +4 | -1
The boiling point elevation constant for toluene is 3.32 K kg mol−13.32 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}. The normal boiling point of toluene is 110.7∘C110.7^{\circ} \mathrm{C}. The enthalpy of vaporisation of toluene would be nearly:
View answer and explanation
Correct option: B
34.0 kJ mol−134.0 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option B (34.0 kJ mol−134.0 \mathrm{~kJ} \mathrm{~mol}^{-1}) is correct because it accurately calculates the enthalpy of vaporization using the boiling point elevation constant and the normal boiling point of toluene.
268 Easy
268 of 460
Application +4 | -1
According to phase rule, if P=3,C=1,P = 3,C = 1, then F must be equal to :
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because, according to the phase rule, F = C - P + 2, and substituting P = 3 and C = 1 gives F = 1 - 3 + 2 = 0.
269 Hard
269 of 460
Analyzing +4 | -1
At 35∘C35^\circ \mathrm{C}, the vapour pressure of CS2C{S_2} is 512 mm Hg and that of acetone is 344 mm Hg. A solution of CS2C{S_2} in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the following is
View answer and explanation
Correct option: C
a mixture of 100 mL CS2C{S_2} and 100 mL acetone has a volume < 200 mL
Option C (a mixture of 100 mL CS2 and 100 mL acetone has a volume < 200 mL) is false because the solution exhibits positive deviation from Raoult's law, meaning the volume of mixing is positive (greater than 200 mL).
270 Easy
270 of 460
Application +4 | -1
A cane sugar solution has an osmotic pressure of 2.46 atm at 300K. What is the strength of the solution.
View answer and explanation
Correct option: A
0.1 M0.1\,M
Option A (0.1 M0.1\,M) is correct because it is calculated using the osmotic pressure formula.
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