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JEE • CHEMISTRY • SOLUTIONS

Solutions practice · Mixed difficulty

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211 Easy
211 of 460
Remembering +4 | -1
For a dilute solution, Raoult's law states that:
View answer and explanation
Correct option: B
the relative lowering of vapour pressure is equal to the mole fraction of solute
Option B is correct because Raoult's law for dilute solutions states that the relative lowering of vapor pressure is equal to the mole fraction of the solute.
212 Easy
212 of 460
Remembering +4 | -1
The solution having lesser value of osmotic pressure is called:
View answer and explanation
Correct option: A
Hypotonic solution
Option A is correct because a hypotonic solution has a lower osmotic pressure compared to another solution.
213 Easy
213 of 460
Application +4 | -1
What is the molarity of 0.2  N  Na2CO30.2\;N\;N{a_2}C{O_3} solution?
View answer and explanation
Correct option: A
0.1 M
Option A (0.1 M) is correct because the molarity is half of the normality for Na2CO3, since it's a dipositive salt.
214 Medium
214 of 460
Understanding +4 | -1
If the various terms in the below given expressions have usual meanings, the van’t Hoff factor (i) cannot be calculated by which one of the expressions?
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Correct option: A
πV=inRT\pi V = \sqrt {inRT}
The correct answer is A (πV=inRT\pi V = \sqrt {inRT}) because the correct van't Hoff equation for osmotic pressure is πV=inRT\pi V = i nRT.
215 Medium
215 of 460
Analyzing +4 | -1
0.1 M NaCl0.1 \mathrm{~M} \mathrm{~NaCl} and 0,005 M BaCl20,005 \mathrm{~M} \mathrm{~BaCl}_2 solutions are separated by a semipermeable membrane in a container. For this system, choose the correct answer.
View answer and explanation
Correct option: B
Water flows from BaCl2\mathrm{BaCl}_2 solution to NaCl\mathrm{NaCl} solution
The correct answer is C because water flows from the BaCl2 solution to the NaCl solution due to the lower solute concentration (higher water potential) in the BaCl2 solution.
216 Easy
216 of 460
Application +4 | -1
In water saturated air, the molefraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is
View answer and explanation
Correct option: C
1.176 atm
Option C (1.176 atm) is correct because the partial pressure of dry air is the total pressure minus the partial pressure of water vapor, calculated using mole fraction.
217 Medium
217 of 460
Application +4 | -1
Find final molarity of solution made by 500 ml 0.1 M HCl+500 ml 0.2 M HCl.500 \mathrm{~ml} \, 0.1 \mathrm{~M} \, {HCl}+500 \mathrm{~ml} \, 0.2 \mathrm{~M} \, {HCl}.
View answer and explanation
Correct option: A
0.15 M
The correct answer is A (0.15 M) because the final molarity is calculated by summing the moles of HCl and dividing by the total volume.
218 Medium
218 of 460
Understanding +4 | -1
Assertion (A): On increasing temperature vapour pressure of solution increases.

Reason (R): Vapour pressure of ether is higher than alcohol.
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Correct option: B
Both A & R are true, R is not correct explanation of A
Option B is correct because both the assertion and the reason are true, but the reason does not explain the assertion. Vapour pressure increases with temperature, and ether has a higher vapour pressure than alcohol, but these are independent facts.
219 Easy
219 of 460
Application +4 | -1
0.1M0.1 {M} solution of glucose was found to be isotonic with a solution of XX in 100 g100 {~g} water. Calculate relative lowering in vapour pressure of solution of XX in water.
View answer and explanation
Correct option: A
0.0018
Option A (0.0018) is correct because it accurately calculates the relative lowering in vapor pressure using the isotonic condition and Raoult's law.
220 Easy
220 of 460
Application +4 | -1
An aqueous solution of glucose containing 12 g12 {~g} in 100 g100 {~g} of water was found to boil at 100.34∘C100.34^{\circ} {C}. Calculate KbK_b for water in Kmol−1 kg{K} {mol}^{-1} {~kg}.
View answer and explanation
Correct option: A
0.51
Option A (0.51) is correct because it represents the ebullioscopic constant (KbK_b) calculated using the elevation in boiling point and the molality of the glucose solution.
221 Easy
221 of 460
Application +4 | -1
Lowering of vapour pressure is highest for
View answer and explanation
Correct option: A
0.1  M  BaCl20.1\;\mathrm{M }\;\mathrm{BaC}{\mathrm{l}_2}
The correct answer is A (0.1  M  BaCl20.1\;\mathrm{M }\;\mathrm{BaC}{\mathrm{l}_2}) because it has the highest van't Hoff factor, leading to the greatest lowering of vapor pressure.
222 Medium
222 of 460
Analyzing +4 | -1
Which of the following will have the lowest vapour pressure
View answer and explanation
Correct option: D
0.1 M K4Fe(CN)60.1\, {M\,K}_4 {Fe}({CN})_6 solution
Option D is correct. 0.1 M K4Fe(CN)60.1\, {M\,K}_4 {Fe}({CN})_6 solution will have the lowest vapor pressure because it dissociates into the highest number of ions (5), resulting in the greatest lowering of vapor pressure due to colligative properties.
223 Medium
223 of 460
Analyzing +4 | -1
Which of the given solutions has highest osmotic pressure?
View answer and explanation
Correct option: A
1  N  NaNO31\;N\;\mathrm{NaN}{\mathrm{O}_3}
Option A is correct because osmotic pressure is directly proportional to the number of particles in the solution. 1NNaNO31N \mathrm{NaNO_3} produces the highest molarity of particles.
224 Medium
224 of 460
Application +4 | -1
A solution of 6%6 \% (mass /vol/ {vol}.) urea is isotonic with NaClaq{NaCl}_{a q}. What is the mass/vol. %\% of NaCl{NaCl} in solution?
View answer and explanation
Correct option: B
2.925%2.925 \%
Option B (2.925 %) is correct because the isotonic condition implies equal osmotic pressures, allowing calculation of the NaCl concentration.
225 Medium
225 of 460
Application +4 | -1
The boiling point of CHCl3{CHCl}_3 was raised by 0.323∘C0.323^{\circ} {C} when 0.5143 g0.5143 {~g} of anthracene was dissolved in 35 gCHCl335 {~g} {CHCl}_3. Calculate the molar mass of anthracene. Kb′K_b^{\prime} for CHCl3=3.9 K mol−1 kg{CHCl}_3=3.9 {~K} {~mol}^{-1} {~kg}.
View answer and explanation
Correct option: A
177.42
The molar mass of anthracene is 177.42 g/mol, which is calculated using the elevation in boiling point formula.
226 Medium
226 of 460
Application +4 | -1
H2S,\mathrm{H}_2 \mathrm{S}, a toxic gas with rotten egg like smell, is used for the quantitative analysis. If the solubility of H2S\mathrm{H}_2 \mathrm{S} in water at STP is 0.2 m,0.2 \mathrm{~m}, then Henry's law constant for H2S\mathrm{H}_2 \mathrm{S} in water at 273 K273 \mathrm{~K} is
View answer and explanation
Correct option: D
2.78×107 Pa2.78 \times 10^7 \mathrm{~Pa}
Option D (2.78×107 Pa2.78 \times 10^7 \mathrm{~Pa}) is correct because it is calculated using Henry's law, relating partial pressure to mole fraction in solution.
227 Medium
227 of 460
Application +4 | -1
Determine the molarity of a 40%40 \% solution of HCl\mathrm{HCl} which has a density of 1.2 g mL−11.2 \mathrm{~g} \mathrm{~mL}^{-1}.
View answer and explanation
Correct option: A
13.15 M13.15 \mathrm{~M}
The correct answer is C (A3B4{A_3}{B_4}) because removing face-centered atoms along one axis alters the A:B ratio in the unit cell.
228 Easy
228 of 460
Understanding +4 | -1
Van’t Hoff factor (i)(i):
View answer and explanation
Correct option: C
i=normal  molecular  massobserved  molecular  massi = \frac{{\mathrm{normal}\;\mathrm{ molecular}\;\mathrm{ mass}}}{{\mathrm{observed }\;\mathrm{molecular}\;\mathrm{ mass}}}
Option C (i=normal  molecular  massobserved  molecular  massi = \frac{{\mathrm{normal}\;\mathrm{ molecular}\;\mathrm{ mass}}}{{\mathrm{observed }\;\mathrm{molecular}\;\mathrm{ mass}}}) is correct because the van't Hoff factor is defined as the ratio of normal molecular mass to observed molecular mass.
229 Medium
229 of 460
Understanding +4 | -1
Concentration terms which are independent of temperature is/are
View answer and explanation
Correct option: A
mole fraction and molality
Option A (mole fraction and molality) is correct because both mole fraction and molality are defined in terms of mass, which is independent of temperature.
230 Medium
230 of 460
Application +4 | -1
How much K2Cr2O7{{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}} (Mol. wt. = 294.19) is required to prepare one litre of 0.1 N solution?
View answer and explanation
Correct option: D
4.903  g4.903\mathrm{ }\;g
Option D (4.903  g4.903\mathrm{ }\;g) is correct because it is the amount of K2Cr2O7 required to prepare one liter of 0.1 N solution, calculated using the equivalent weight and normality.
231 Easy
231 of 460
Understanding +4 | -1
Statement I: Raoult’s law is applicable for dilute solutions only.
Statement II: Henry’s law is applicable for solution of gas in liquid.
View answer and explanation
Correct option: A
Statement I and statement II both are correct
Option A (Statement I and statement II both are correct) is correct because Raoult's law is indeed applicable for dilute solutions only, and Henry's law applies to the solution of gas in liquid.
232 Medium
232 of 460
Application +4 | -1
At 100∘C,100^{\circ} \mathrm{C}, the vapour pressure of a solution of 4.0 g4.0 \mathrm{~g} of solute in 100 g100 \mathrm{~g} of water is 750 mm.750 \mathrm{~mm}. The boiling point of the solution is (Kb\left(\mathrm{K}_{\mathrm{b}}\right. of water =0.52=0.52 K−kg/mol)\mathrm{K}-\mathrm{kg} / \mathrm{mol})
View answer and explanation
Correct option: C
100.4∘C100.4^{\circ} \mathrm{C}
Option C (100.4°C) is correct because the boiling point elevation is calculated using the molality of the solution and the ebullioscopic constant.
233 Easy
233 of 460
Application +4 | -1
The osmotic pressure of 0.1 M0.1 \,{M} aq. glucose solution at 27∘C27^{\circ} {C} is
View answer and explanation
Correct option: B
2.46 atm2.46 {~atm}
The correct answer is B (2.46 atm2.46 {~atm}) because it correctly applies the osmotic pressure formula using the given concentration, gas constant, and temperature.
234 Hard
234 of 460
Application +4 | -1
An organic liquid, AA is immiscible with water. When boiled together with water, the boiling point is 90∘C90^{\circ} \mathrm{C} at which the partial pressure of water is 526 mmHg526 \mathrm{~mm} \mathrm{Hg}. The atmospheric pressure is 736 mmHg736 \mathrm{~mm} \mathrm{Hg}. The weight ratio of the liquid and water collected is 2.5:12.5: 1. The molecular weight of the liquid is ...
View answer and explanation
Correct option: A
112.7
Option A (112.7) is correct because the molecular weight of the liquid is calculated using the vapor pressure and weight ratio of the liquid and water.
235 Medium
235 of 460
Application +4 | -1
The solubility of iodine in water is 0.8  g/L{0.8\;g/L}. If the partition coefficient of iodine between CCl4{\mathrm{CC}{\mathrm{l}_4}} and water (in favour of CCl4{\mathrm{CC}{\mathrm{l}_4}}) is 82, the solubility of iodine in CCl4{\mathrm{CC}{\mathrm{l}_4}} is :
View answer and explanation
Correct option: B
65.6  g/L65.6\mathrm{ }\;g/L
Option B (65.6 g/L) is correct because the solubility in CCl4 is calculated using the partition coefficient and the solubility in water.
236 Medium
236 of 460
Analyzing +4 | -1
Assertion: Van't Hoff factor for benzoic acid in benzene is one.

Reason: Benzoic acid behaves as a weaker electrolyte in benzene.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because benzoic acid dimerizes in benzene, leading to a Van't Hoff factor close to 0.5, not 1, and it doesn't behave as a weaker electrolyte.
237 Easy
237 of 460
Remembering +4 | -1
Relative lowering in vapour pressure of solution containing non-volatile solute is directly proportional to mole fraction of solute". Above statement is:
View answer and explanation
Correct option: C
Raoult's law
Option C (Raoult's law) is correct because Raoult's law states that the relative lowering of vapor pressure of a solution containing a non-volatile solute is directly proportional to the mole fraction of the solute.
238 Medium
238 of 460
Analyzing +4 | -1
Assertion : Themolecular weight of acetic acid determined by depression in freezing point method in benzene and water was found to be different.

Reason : Water is polar and benzene is non-polar.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
Option A is correct because both the assertion and reason are true, and the reason correctly explains why the molecular weight of acetic acid differs in benzene and water due to differences in polarity and association.
239 Hard
239 of 460
Application +4 | -1
The boiling point of water (100∘C)\left(100^{\circ} {C}\right) becomes 100.52∘C100.52^{\circ} {C}, if 3 grams3 \,grams of a nonvolatile solute is dissolved in 200 ml200 \,{ml} of water. The molecular weight of solute is ( KbK_b for water is 0.6 K−m0.6 {~K}-{m} )
View answer and explanation
Correct option: C
17.3 g mol−117.3 \,{g\,mol}^{-1}
Option C (17.3 g/mol⁻¹) is correct because it is calculated using the colligative property formula for elevation in boiling point.
240 Easy
240 of 460
Application +4 | -1
An aqueous solution of NaCl{NaCl} shows the depression of freezing point of water equal to 0.372 K0.372 {~K}. The boiling point of BaCl2{BaCl}_2 solution of same molality will be (Kf(H2O)=1.86 K kg mol−1\left(K_f\left({H}_2 {O}\right)=1.86 {~K} {~kg} {~mol}^{-1}\right., Kb(H2O)=0.52 K kg mol−1)\left.K_b\left({H}_2 {O}\right)=0.52 {~K} {~kg} {~mol}^{-1}\right)
View answer and explanation
Correct option: D
100.156∘C100.156^{\circ} {C}
Option D (100.156∘C100.156^{\circ} {C}) is correct because it involves calculating the elevation in boiling point using colligative properties and van't Hoff factors for NaCl and BaCl2.
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