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JEE • CHEMISTRY • SOLUTIONS

Solutions practice · Mixed difficulty

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151 Medium
151 of 460
Application +4 | -1
The correct option for the value of vapour pressure of a solution at 450C{45^0}C with benzene to octane in molar ratio 3:23:2 is
[At 450C{45^0}C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume ideal gas]
View answer and explanation
Correct option: A
336 mm of Hg
Option A (336 mm of Hg) is correct because it calculates the vapor pressure of the solution using Raoult's Law, assuming ideal behavior.
152 Hard
152 of 460
Analyzing +4 | -1
Liquids 'A' and 'B' form an ideal solution. At 80∘C,PA0=0.4 bar80^{\circ} \mathrm{C}, \mathrm{P}_{\mathrm{A}}^0=0.4 \mathrm{~bar} and PB0=0.8 bar.\mathrm{P}_{\mathrm{B}}^0=0.8 \mathrm{~bar}. All the vapour above the liquid solution containing equal moles of both the liquids at equilibrium is collected in another empty vessel and condensed. Now, the condesate is heated to 80∘C80^{\circ} \mathrm{C} and all the vapours above the liquid solution at equilibrium is again collected in another empty vessel and condensed. What is the mole-fraction of 'B' is new condensate?
View answer and explanation
Correct option: A
0.8
The correct answer is A (0.8) because after two condensation cycles, the mole fraction of B is enriched due to its higher vapor pressure.
153 Medium
153 of 460
Application +4 | -1
Which will have largest ΔTb?\Delta {T}_{{b}}?
View answer and explanation
Correct option: D
65 g of urea in 1 kg of water
Option D (65 g of urea in 1 kg of water) is correct because urea is a non-electrolyte, and this solution has the highest molality compared to the other options, leading to the largest boiling point elevation.
154 Medium
154 of 460
Understanding +4 | -1
Which relation is incorrect for a solution that shows positive deviation from Raoult's law?
View answer and explanation
Correct option: D
(ΔV)mix <0(\Delta V)_{\text {mix }} < 0
The correct answer is D ((ΔV)mix <0(\Delta V)_{\text {mix }} < 0) because solutions with positive deviation from Raoult's law have a positive volume change upon mixing.
155 Easy
155 of 460
Application +4 | -1
The vapour pressure of 100 g100 \mathrm{~g} water reduces from 17.53 mm17.53 \mathrm{~mm} to 17.22 mm17.22 \mathrm{~mm} when 17.10 g17.10 \mathrm{~g} substance ' XX ' is dissolved in it. Substance XX can be
View answer and explanation
Correct option: B
glucose
Option B (glucose) is correct because the calculated molar mass of the unknown substance X is approximately 174 g/mol, which is close to the molar mass of glucose.
156 Medium
156 of 460
Application +4 | -1
A solution containing 8.6 g8.6 \mathrm{~g} urea in one litre was found to be isotonic with a 5%5 \% (mass/volume) solution of an organic non-volatile solute. The molar mass of solute is;
View answer and explanation
Correct option: A
348.83
Option A (348.83) is correct because it is the molar mass of the solute calculated using the isotonic condition and the given concentrations.
157 Medium
157 of 460
Application +4 | -1
What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml100 \,{ml} of 0.2 N−NaOH0.2 {~N}-{NaOH} solution?
View answer and explanation
Correct option: D
1.26 g1.26 {~g}
Option D (1.26 g1.26 {~g}) is correct because it is the mass of hydrated oxalic acid required for complete neutralization, calculated using the concept of equivalents.
158 Medium
158 of 460
Application +4 | -1
On dissolving 0.5 g0.5 {~g} of a non-volatile non-ionic solute to 39 g39 {~g} of benzene, its vapour pressure decreases from 650 mmHg650 {~mm} {Hg} to 640 mmHg640 {~mm} {Hg}. The depression of freezing point of benzene (in K) upon addition of the solute is..... K.
(Given data : Molar mass and the molal freezing point depression constant of benzene are 78 g mol−178 {~g} {~mol}^{-1} and 5.12 K kg mol−15.12 {~K} {~kg} {~mol}^{-1}, respectively)
View answer and explanation
Correct option: A
1.03
Option A (1.03) is correct because it's the freezing point depression calculated using the colligative properties formula and the given data.
159 Medium
159 of 460
Analyzing +4 | -1
Which of the following mixture of solutions follow the condition Δmix H<0\Delta_{\text {mix }} H<0 and Δmix V<0\Delta_{\text {mix }} V<0 ?

(i) Nitric acid + Water

(ii) Acetic acid + Pyridine

(iii) Ethanol + Water

(iv) Carbon tetra chloride + Chloroform
View answer and explanation
Correct option: D
(i) & (ii) only
The correct answer is D ((i) & (ii) only) because nitric acid + water and acetic acid + pyridine mixtures exhibit negative enthalpy and volume changes upon mixing.
160 Medium
160 of 460
Application +4 | -1
KspK_{s p} (solubility product) of MgS=(Mg2+(S2−))\mathrm{MgS}=\left(\mathrm{Mg}^{2+}\left(\mathrm{S}^{2-}\right)\right) is 1.0×10−61.0 \times 10^{-6}. Thus, elevation in boiling point of saturated solution is (Kb=0.52∘mol−1 kg)\left(K_b=0.52^{\circ} \mathrm{mol}^{-1} \mathrm{~kg}\right)
View answer and explanation
Correct option: D
5.2×10−4 K5.2 \times 10^{-4} \mathrm{~K}
Option D (5.2×10−4 K5.2 \times 10^{-4} \mathrm{~K}) is correct because it's calculated by finding the solubility from Ksp and then using the elevation in boiling point formula.
161 Easy
161 of 460
Understanding +4 | -1
The statement “the relative lowering of the vapour pressure is equal to th ratio ot moles of the solute to the total number of the moles in the solution” refers to
View answer and explanation
Correct option: A
Decrease the freezing point of water in the winter and increase the boiling point of water in the summer
The correct answer is A because the statement describes colligative properties, specifically the effect of solutes on freezing and boiling points.
162 Medium
162 of 460
Remembering +4 | -1
Among the following, the azeotropic mixture is
View answer and explanation
Correct option: A
CCl4+CHCl3\mathrm{CCl}_4+\mathrm{CHCl}_3
Option A (CCl4+CHCl3\mathrm{CCl}_4+\mathrm{CHCl}_3) is correct because CCl4\mathrm{CCl}_4 and CHCl3\mathrm{CHCl}_3 form a non-ideal solution and can exhibit azeotropic behavior.
163 Easy
163 of 460
Understanding +4 | -1
A liquid is kept in a closed vessel. If a glass plate (negligible mass) with a small hole is kept on top of the liquid surface, then the vapour pressure of the liquid in the vessel is :
View answer and explanation
Correct option: B
Same as what would be if the glass plate were removed
The correct answer is B (Same as what would be if the glass plate were removed) because vapor pressure depends only on temperature and the nature of the liquid, not the presence of a glass plate.
164 Medium
164 of 460
Analyzing +4 | -1
Assertion : Camphor is used as solvent in the determination of molecular masses of naphthalene, anthracene etc.

Reason : Camphor has high molal elevation constant.
View answer and explanation
Correct option: C
If assertion is true but reason is false.
The correct answer is C because camphor is indeed used as a solvent for molecular mass determination due to its high molal depression constant, not elevation constant.
165 Easy
165 of 460
Remembering +4 | -1
The relative lowering of vapour pressure is equal to the mole fraction of the non-volatile solute. This statement was given by:
View answer and explanation
Correct option: A
Raoult
Option A (Raoult) is correct because Raoult's Law states that the relative lowering of vapor pressure is equal to the mole fraction of the non-volatile solute.
166 Medium
166 of 460
Analyzing +4 | -1
Consider the following statements about osmotic pressure method of molar mass determination.

I. Molarity of the solution is used instead of molality.

II. Magnitude of osmotic pressure is very large even for dilute solutions.

III. Molar mass of biomolecules can be determined as they are not stable at higher temperatures.

IV. Determine the molar mass of polymer as they have poor solubility.

Which of the above statements are responsible for advantage of osmotic pressure for determination of molar mass over the other colligative properties?
View answer and explanation
Correct option: C
I, II, III and IV
Option C (I, II, III and IV) is correct because all the listed statements contribute to the advantages of using osmotic pressure for molar mass determination.
167 Medium
167 of 460
Analyzing +4 | -1
Assertion (A) : Addition of solvent to a solution always lowers the V.P.

Reason (R) : The increase in relative surface area give rise to an increase in V.P. for a given solution.
View answer and explanation
Correct option: B
Assertion is Wrong but Reason is correct
Option B is correct. Adding a solvent to a solution *increases* the vapor pressure, not lowers it. The reason is correct because increasing the relative surface area allows more solvent molecules to escape into the vapor phase.
168 Medium
168 of 460
Application +4 | -1
The vapour pressure of acetone at 20∘C20^\circ C is 185 torr. When 1.2 g of a non – volatile substance was dissolved in 100 g of acetone at 20∘C,20^\circ C, its vapour pressure was 183 torr. The molar mass (g  mol−1)(\mathrm{g}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}) of the substance is
View answer and explanation
Correct option: B
64
The correct answer is 64 because the molar mass of the non-volatile substance is approximately 64 g/mol, calculated using Raoult's Law for vapor pressure lowering.
169 Easy
169 of 460
Remembering +4 | -1
According to Raoult's law, relative lowering of vapour pressure for a solution is equal to
View answer and explanation
Correct option: C
Mole fraction of solute
Option C (Mole fraction of solute) is correct because Raoult's law states that the relative lowering of vapor pressure is equal to the mole fraction of the solute in the solution.
170 Medium
170 of 460
Analyzing +4 | -1
Assertion: The molarity and normality of a solution of sodium carbonate are same.
Reason: Molarity is the product of normality &amp; valency factor.
View answer and explanation
Correct option: D
If both Assertion &amp; Reason are false
The correct answer is If both Assertion & Reason are false because the molarity and normality of sodium carbonate are different, and molarity is not the product of normality and valency factor.
171 Medium
171 of 460
Application +4 | -1
How much water is needed to dilute 10 ml10 {~ml} of 10 N10 {~N} hydrochloric acid to make it exactly decinormal (0.1 N)(0.1 {~N})
View answer and explanation
Correct option: A
990 ml990 \,{ml}
Option A (990 mL) is correct because it is calculated using the dilution formula N1V1=N2V2N_1V_1 = N_2V_2 and then subtracting the initial volume.
172 Medium
172 of 460
Application +4 | -1
Vapour pressure of pure water at 298 K298 \mathrm{~K} is 23.8 mm23.8 \mathrm{~mm}. Thus, vapour pressure of 20%20 \% glucose (weight/volume) solution is (d=0.9971 g mL−1)\left(d=0.9971 \mathrm{~g} \mathrm{~mL}^{-1}\right)
View answer and explanation
Correct option: D
23.217 mm23.217 \mathrm{~mm}
Option D (23.217 mm23.217 \mathrm{~mm}) is correct because it uses Raoult's law to accurately calculate the vapor pressure lowering of the solution.
173 Medium
173 of 460
Application +4 | -1
The freezing point of a solution containing 0.2 g0.2 \mathrm{~g} of acetic acid in 20.0 g20.0 \mathrm{~g} benzene is lowered by 0.45∘C0.45^{\circ} \mathrm{C}. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and Kf\mathrm{K}_{\mathrm{f}} for benzene =5.12 K kg mol−1=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1})
Mobserved\mathrm{M}_{\text {observed}} of acetic acid =113.78=113.78
View answer and explanation
Correct option: A
94.5%94.5 \%
Option A (94.5%94.5 \%) is correct because the degree of association is calculated using the van't Hoff factor and the observed molar mass.
174 Medium
174 of 460
Application +4 | -1
The amount of K2Cr2O7{K}_2 {Cr}_2 {O}_7 (eq. wt. 49.04) required to prepare 100 ml {ml}\, of its 0.05 N0.05 {~N} solution is
View answer and explanation
Correct option: D
0.2452 g0.2452 {~g}
Option D (0.2452 g0.2452 {~g}) is correct because it is the amount of K2Cr2O7 required to prepare 100 ml of 0.05 N solution, calculated using the normality equation.
175 Medium
175 of 460
Analyzing +4 | -1
The amino acid alanine has two isomers, such as
α−alanine\alpha-\text{alanine} and β−alanine.\beta-\text{alanine}. When equal masses of
these two compounds are dissolved in equal mass
of a solvent, the solution of α−alanine\alpha-\text{alanine} freezes at
relatively lower temperature. Which forms of
α−alanine\alpha-\text{alanine} or β−alanine\beta-\text{alanine}, has the larger equilibrium
constant for ionization?
View answer and explanation
Correct option: A
α−alanine\alpha-\text{alanine}
Option A (α−alanine\alpha-\text{alanine}) is correct because a lower freezing point indicates a greater number of particles in solution, implying a larger equilibrium constant for ionization.
176 Medium
176 of 460
Analyzing +4 | -1
S-I: The boiling point of 0.1 M0.1 \, \mathrm{M} urea solution is less than that of 0.1 M KCl0.1 \, \mathrm{M} \, {KCl} solution.

S-II: Elevation of boiling point is directly proportional to the number of moles of non-volatile solute particles present in the solution.
View answer and explanation
Correct option: A
Both S-I and S-II are correct
Option A (Both S-I and S-II are correct) is correct because the boiling point elevation depends on the number of solute particles, and KCl dissociates into more particles than urea.
177 Medium
177 of 460
Application +4 | -1
The molarity of 24.5%24.5 \% (w/w) H2SO4\mathrm{H}_2 \mathrm{SO}_4 solution having density 1.2 g/ml1.2 \mathrm{~g} / \mathrm{ml} is
View answer and explanation
Correct option: B
3 M3 \mathrm{~M}
Option B (3 M) is correct because molarity is calculated using the formula Molarity = (x * d * 10) / M, where x is the percentage by weight, d is the density, and M is the molecular weight.
178 Medium
178 of 460
Application +4 | -1
Which of the following aqueous solution has minimum freezing point?
View answer and explanation
Correct option: B
0.05 m0.05 \mathrm{~m} glucose
Option B (0.05 m0.05 \mathrm{~m} glucose) is correct because a higher effective molality of solute results in a lower freezing point, and glucose does not dissociate.
179 Medium
179 of 460
Application +4 | -1
If the mass percentage (w/w)({w} / {w}) of urea in aqueous solution is 18%18 \% then the molality of the solution is
View answer and explanation
Correct option: D
3.66  m3.66\, {~m}
Option D (3.66 m) is correct because it is the molality calculated from the given mass percentage of urea in the aqueous solution.
180 Medium
180 of 460
Application +4 | -1
Percentage ionization of K3[Fe(CN)6]{K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right] whose decimolar solution shows an osmotic pressure of 3.94 atm at 300 K is
View answer and explanation
Correct option: C
0.2
Option C (0.2) is incorrect, the correct answer is 20% ionization. The percentage ionization is calculated using the osmotic pressure and Van't Hoff factor.
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