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JEE • CHEMISTRY • SOLUTIONS

Solutions practice · Medium

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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 31-60 of 265 questions
31 Medium
31 of 265
Analyzing +4 | -1
1.0 molar solution of the complex salt, CrCl3⋅6H2O\mathrm{CrCl}_3 \cdot 6 \mathrm{H}_2 \mathrm{O}, displays an osmotic pressure of 3RT3 R T. 0.5 L0.5 \mathrm{~L} of the same solution on treatment with excess of AgNO3\mathrm{AgNO}_3 solution will yield (assume α=1\alpha=1 ) :
View answer and explanation
Correct option: B
1.0 mole of AgCl1.0 \text { mole of } \mathrm{AgCl}
Option B (1.0 mole of AgCl1.0 \text{{ mole of }} \mathrm{AgCl}) is correct because the complex is [Cr(H2O)5Cl]Cl2⋅H2O\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}, and only the chloride ions outside the coordination sphere react with AgNO3AgNO_3.
32 Medium
32 of 265
Application +4 | -1
Calculate the amount of ice that will separate out on cooling aa solution containing 50 g50 {~g} of ethylene glycol in 200 g200 {~g} water to −9.3∘C-9.3^{\circ} {C}. ( KfK_f for water =1.86 K mol−1 kg=1.86 {~K} {~mol}^{-1} {~kg} )
View answer and explanation
Correct option: A
38.71
Option A (38.71) is correct because it involves calculating the amount of water remaining in the solution at -9.3°C using the freezing point depression formula and then subtracting that from the initial amount of water to find the mass of ice separated.
33 Medium
33 of 265
Analyzing +4 | -1
During which of the following processes, does entropy decrease?
A) Freezing of water to ice at 0∘C0^\circ C.
B) Freezing of water to ice at −10∘C- 10^\circ C.
C) N2(g)+3H2(g)→2NH3(g){N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)
D) Adsorption of CO(g)CO(g) and lead surface.
E) Dissolution of NaClNaCl in water.
View answer and explanation
Correct option: A
A, B, C and D
Option A is correct because freezing, gas-to-solid conversion, and adsorption all result in a decrease in entropy due to increased order.
34 Medium
34 of 265
Application +4 | -1
A solution of urea boils at 100.180C{100.18^0}C at the atmospheric pressure. If Kf{K_f} and Kb{K_b} for water are 1.86 and 0.512 K.Kg.Mol−1K.Kg.Mo{l^{ - 1}} respectively, the above solution will freeze at
View answer and explanation
Correct option: B
−0.654 0C- 0.654{\,^0}C
Option B (-0.654 °C) is correct because the freezing point depression can be calculated using the boiling point elevation and the cryoscopic and ebullioscopic constants.
35 Medium
35 of 265
Application +4 | -1
An aqueous solution of a weak monobasic acid containing 0.1g0.1 g in 21.7g21.7 g of water freezes at 272.813 K272.813\, K. If the value of KfK_f for water is 1.86 K/m1.86 {~K} / {m}, what is the molecular mass of the monobic acid
View answer and explanation
Correct option: D
60 g/mole60 {~g} / {mole}
The correct answer is D (60 g/mole) because the molecular mass is calculated using the freezing point depression formula.
36 Medium
36 of 265
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.
37 Medium
37 of 265
Understanding +4 | -1
Assertion (A): People taking a lot of salty food experience the puffiness or swelling, called edema.

Reason (R): There is water retention in tissue cells and intercellular spaces hecause of osmosis.
View answer and explanation
Correct option: A
Both A\mathrm{A} and R\mathrm{R} are correct, R\mathrm{R} is the correct explanation of A\mathrm{A}.
Option A (Both A and R are correct, R is the correct explanation of A) is correct because salty food increases the solute concentration in the body, causing water retention in tissues due to osmosis, leading to edema.
38 Medium
38 of 265
Application +4 | -1
Consider the following arrangement, in which a solution containing 20 g20 \mathrm{~g} of haemoglobin in 1 dm31 \mathrm{~dm}^3 of the solution is placed in right compartment and pure water is placed in left compartment, separated by SPM. At equilibrium, the height of liquid in the right compartment is 74.5 mm74.5 \mathrm{~mm} in excess of that in the left compartment. The temperature of the system is maintained at 298 K.298 \mathrm{~K}. The number of millimoles in 320 g320 \mathrm{~g} of haemoglobin is (Given: The density of final solution is 1.013 g/ml,g=10 m/s21.013 \mathrm{~g} / \mathrm{ml}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, R=0.08 L−atm/K−molR=0.08 \mathrm{~L}-\mathrm{atm} / \mathrm{K}-\mathrm{mol})
View answer and explanation
Correct option: A
5
Option A (5) is correct because the number of millimoles is calculated using the osmotic pressure equation and the given parameters.
39 Medium
39 of 265
Application +4 | -1
If the elevation in boiling point of a solution of 10 g of solute (mol. wt. = 100 g of water is ΔTb\Delta {T_b}, the ebullioscopic constant of water is
View answer and explanation
Correct option: C
ΔTb\Delta {T_b}
Option C (ΔTb\Delta {T_b}) is correct because the ebullioscopic constant is equal to the elevation in boiling point under the given conditions.
40 Medium
40 of 265
Application +4 | -1
2.5 g2.5 {~g} of a substance is present in 200 mL200 {~mL} of solution showing the osmotic pressure of 60 cmHg60 {~cm} {Hg} at 15∘C15^{\circ} {C}. Calculate the molar mass of substance. What will be the osmotic pressure if temperature is raised to 25∘C25^{\circ} {C} ?
View answer and explanation
Correct option: A
374.38
Option A (374.38) is correct because it is calculated using the osmotic pressure formula and the ideal gas constant.
41 Medium
41 of 265
Analyzing +4 | -1
Each of the following pair shows a positive deviation from the Raoult's law except
View answer and explanation
Correct option: A
(C2H5)O+HCl\left({C}_2 {H}_5\right) {O}+{HCl}
Option A ((C2H5)O+HCl\left({C}_2 {H}_5\right) {O}+{HCl}) is correct because ether and HCl form hydrogen bonds, leading to stronger interactions and a negative deviation from Raoult's law.
42 Medium
42 of 265
Understanding +4 | -1
Consider the solution of ethanol. In it
View answer and explanation
Correct option: D
Both (1) and (2)
Option D (Both (1) and (2)) is correct because in a solution of ethanol, ethanol is dissolved in water and both components are in the same phase (liquid).
43 Medium
43 of 265
Analyzing +4 | -1
Select correct statements :
View answer and explanation
Correct option: D
All are correct statements
The correct answer is D (All are correct statements) because all the statements regarding colligative properties, freezing point depression, and isotonic solutions are accurate.
44 Medium
44 of 265
Analyzing +4 | -1
Statement I: Mass density of a mixture of 20 g20 \mathrm{~g} water and 100 g100 \mathrm{~g} of ethanol is 0.86 g cm−30.86 \mathrm{~g} \mathrm{~cm}^{-3} at 298 K298 \mathrm{~K}.

Statement II: At 298 K298 \mathrm{~K}, molar volume of water and ethanol are respectively 18 cm318 \mathrm{~cm}^3 and 55 cm355 \mathrm{~cm}^3 based on density. (d=1.00 g cm−3\left(d=1.00 \mathrm{~g} \mathrm{~cm}^{-3}\right. for water and d=0.8364 g cm−3d=0.8364 \mathrm{~g} \mathrm{~cm}^{-3} for ethanol).
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I.
Option B (Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I) is correct because both statements are factually correct, but the molar volumes do not directly explain the mixture's density.
45 Medium
45 of 265
Understanding +4 | -1
Statement I: Normality and molarity can be calculated from each other.

Statement II: Normality is equal to the product of molarity and n\mathrm{n}.
View answer and explanation
Correct option: A
If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
Option A is correct because Normality is defined as the product of Molarity and the n-factor (number of equivalents), making the statements correct and related.
46 Medium
46 of 265
Application +4 | -1
An aqueous solution of a non-volatile and non-electrolyte solute (molecular mass = 150) boils at 373.26 K.373.26 \mathrm{~K}. The composition of solution, in terms of mass percent of the solute, is (Kb\left(K_{\mathrm{b}}\right. of water == 0.52)0.52)
View answer and explanation
Correct option: C
6.98%6.98 \%
Option C (6.98 %) is correct because it uses the elevation in boiling point formula to find the molality and then calculates the mass percent of the solute.
47 Medium
47 of 265
Application +4 | -1
Which of the following has lowest freezing point
View answer and explanation
Correct option: B
0.1 M0.1 \,{M} aqueous solution of NaCl{NaCl}
Option B (5.6 V) is correct because the volume strength of 1N H2O2 is 5.6 V, which is a direct application of the formula.
48 Medium
48 of 265
Understanding +4 | -1
A perfectly semipermeable membrane when used to separate a solution from its solvent permits through it the passage of:
View answer and explanation
Correct option: B
Solvent only
Option B (Solvent only) is correct because a perfectly semipermeable membrane allows only solvent molecules to pass through, blocking solute molecules.
49 Medium
49 of 265
Understanding +4 | -1
At higher altitudes the boiling point of water lowers because
View answer and explanation
Correct option: A
Atmospheric pressure is low
Option A is correct because at higher altitudes, the atmospheric pressure is lower, which causes water to boil at a lower temperature.
50 Medium
50 of 265
Understanding +4 | -1
Which one of the following gases has the lowest value of Henry's law constant?
View answer and explanation
Correct option: D
CO2\mathrm{CO}_2
Option D (CO2\mathrm{CO}_2) is correct because carbon dioxide has the highest solubility in water among the given gases, leading to the lowest Henry's law constant.
51 Medium
51 of 265
Analyzing +4 | -1
An open beaker of water in equilibrium with water vapour is in a sealed container. When few grams of glucose are added to the beaker of water, the rate at which water molecules
View answer and explanation
Correct option: C
leaves the vapour increases
Option C is correct because adding glucose to water lowers the water's vapor pressure, causing more water to evaporate to re-establish equilibrium.
52 Medium
52 of 265
Application +4 | -1
One mole of a solute A\mathrm{A} is dissolved in a given volume of a solvent. The association of the solute take place as follows: nA⇌An\mathrm{nA} \rightleftharpoons \mathrm{A}_{\mathrm{n}}. If α\alpha is the degree of association of A, the van't Hoff factor ' i\mathrm{i} ' is expressed as:
View answer and explanation
Correct option: C
i=1−α+αn1\mathrm{i}=\frac{1-\alpha+\frac{\alpha}{\mathrm{n}}}{1}
Option C (6 min) is correct because it is the half-life calculated from the rate constants at two different times for a first-order reaction.
53 Medium
53 of 265
Analyzing +4 | -1
Assertion (A): An azeotropic solution of two liquids has boiling point lower than either of them.
Reason (R): Solution shows +ve deviation from ideal behaviour.
View answer and explanation
Correct option: A
Both A & R are true, R is the correct explanation of A
Option A (Both A & R are true, R is the correct explanation of A) is correct because azeotropic solutions with lower boiling points exhibit positive deviations from ideal behavior.
54 Medium
54 of 265
Analyzing +4 | -1
Which of the following aqueous solution has osmotic pressure nearest to that of an equimolar solution of K4[Fe(CN)6]?\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] ?
View answer and explanation
Correct option: C
Al2(SO4)3\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3
Option C (Al2(SO4)3\mathrm{Al}_2(\mathrm{SO}_4)_3) is correct because it has a van't Hoff factor (i=5) closest to that of K4[Fe(CN)6]\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6] (i=5), resulting in similar osmotic pressure.
55 Medium
55 of 265
Application +4 | -1
The volume of 0.5 M0.5 \mathrm{~M} Urea required to prepare 0.1 M0.1 \mathrm{~M} of 100 ml100 \mathrm{~ml} Urea solution is
View answer and explanation
Correct option: A
20 ml20 \mathrm{~ml}
Option A (20 ml20 \mathrm{~ml}) is correct because using the dilution formula M1V1=M2V2M_1V_1 = M_2V_2, we find that 20 ml of 0.5 M Urea is required to prepare 100 ml of 0.1 M Urea solution.
56 Medium
56 of 265
Analyzing +4 | -1
Which one of the following statements is false ?
View answer and explanation
Correct option: D
Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression
Option D is correct because the freezing point depression depends on the solvent's cryoscopic constant, which varies between different solvents.
57 Medium
57 of 265
Analyzing +4 | -1
Statement I: Mercuric iodide dissolves in potassium iodide solution forming K2Hgl4\mathrm{K}_2 \mathrm{Hgl}_4 and the freezing point gets elevated.

Statement II: The number of particles in K2Hgl4\mathrm{K}_2 \mathrm{Hgl}_4 is same as those in Hgl2\mathrm{Hgl}_2.
View answer and explanation
Correct option: C
If Statement I is correct but Statement II is false
Option C is correct because Statement I is correct, but Statement II is false because the number of particles increases upon dissolution.
58 Medium
58 of 265
Application +4 | -1
If vapour pressure of two liquids A and B in pure state are 400 and 500 mm of Hg respectively then mole fraction of A in vapour phase of solution obtained by mixing equal moles of A and B is
View answer and explanation
Correct option: C
49\frac{4}{9}
Option C (4/9) is correct because it correctly applies Raoult's Law and Dalton's Law to determine the mole fraction of A in the vapor phase.
59 Medium
59 of 265
Application +4 | -1
A solution containing 8.6 g8.6 \mathrm{~g} urea in one liter was found to be isotonic with a 5%(w/V)5 \%(w / V) solution of an organic non-volatile solute at same temperature. The molecular weight of latter is
View answer and explanation
Correct option: A
348.8 g mol−1348.8 \mathrm{~g} \mathrm{~mol}^{-1}
Option A (348.8 g mol⁻¹) is correct because the molecular weight of the organic solute is calculated using the isotonic condition and the molar mass of urea.
60 Medium
60 of 265
Application +4 | -1
Calculate the boiling point of a one molar aqueous solution (density 1.04 g mL−11.04 \mathrm{~g} \mathrm{~mL}^{-1}) of potassium chloride. (Kb\left(\mathrm{K}_{\mathrm{b}}\right. for water =0.52 K kg mol−1=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}. Atomic mass of K=39,Cl=35.5\mathrm{K}=39, \mathrm{Cl}=35.5)
View answer and explanation
Correct option: A
374.1128 K374.1128 \mathrm{~K}
Option A (374.1128 K374.1128 \mathrm{~K}) is correct because it is calculated by determining the molality, van't Hoff factor, and boiling point elevation, then adding the elevation to the normal boiling point of water.
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