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JEE • CHEMISTRY • SOLUTIONS

Solutions practice · Medium

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151 Medium
151 of 265
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.
152 Medium
152 of 265
Understanding +4 | -1
Which of the following statements is incorrect?
View answer and explanation
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.
153 Medium
153 of 265
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.
154 Medium
154 of 265
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.
155 Medium
155 of 265
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.
156 Medium
156 of 265
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.
157 Medium
157 of 265
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.
158 Medium
158 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.
159 Medium
159 of 265
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.
160 Medium
160 of 265
Application +4 | -1
If 0.2 m aqueous solution of K4[Fe(CN)6]{K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] is 100% ionized then the freezing point of the solution will be (Kf=1.86 K kg mol−1)\left( {{K_f} = 1.86\,K\,kg\,mo{l^{ - 1}}} \right)
View answer and explanation
Correct option: B
−1.860C- {1.86^0}C
Option A (−1.860C- {1.86^0}C) is correct because it's the freezing point calculated using the van't Hoff factor and the freezing point depression formula.
161 Medium
161 of 265
Understanding +4 | -1
Given below are two statements:

Statement (I): NaCl is added to the ice at 0∘C,0^{\circ} \mathrm{C}, present in the ice cream box to prevent the melting of ice cream.

Statement (II): On addition of NaCl to ice at 0∘C,0^{\circ} \mathrm{C}, there is a depression in freezing point.

In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: B
Both Statement I and Statement II are true
The correct answer is B (Both Statement I and Statement II are true) because adding NaCl to ice lowers the melting point, which is the principle behind freezing mixtures.
162 Medium
162 of 265
Analyzing +4 | -1
Assertion: The mixing of two completely miscible liquids. A and B showing positive deviation from Raoult's law is followed by an absorption of heat

Reason: The A-B molecular interaction forces are stronger than the A-A or B-B molecular interaction forces.
View answer and explanation
Correct option: C
Assertion is True, Reason is False
Option C (Assertion is True, Reason is False) is correct because positive deviation implies weaker A-B interactions, contradicting the reason.
163 Medium
163 of 265
Analyzing +4 | -1
Select correct statement ?
View answer and explanation
Correct option: D
All are correct statements
Option D (All are correct statements) is correct because heats of vaporization are similar, entropy change is smaller for solutions, and boiling point elevation occurs in solutions.
164 Medium
164 of 265
Understanding +4 | -1
Which is an application of Henry's law?
View answer and explanation
Correct option: D
Soft drinks (soda)
Option D (Soft drinks (soda)) is correct because Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid, which is applied in carbonating soft drinks.
165 Medium
165 of 265
Understanding +4 | -1
Assertion: After adding non-volatile solute freezing point of solvent depressed.

Reason: Vapour pressure of solvent decreases and become equal to solid solvent, at temperature lower than the freezing point of pure solvent.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the depression in freezing point is directly caused by the lowering of vapor pressure due to the non-volatile solute.
166 Medium
166 of 265
Analyzing +4 | -1
Which has maximum freezing point ?
View answer and explanation
Correct option: A
6 g6 \mathrm{~g} urea solution in 100 g H2O100 \mathrm{~g} \mathrm{~H}_2 \mathrm{O}
The correct answer is A because urea is a non-electrolyte and has the lowest van't Hoff factor (i=1), leading to the least depression in freezing point and thus the highest freezing point.
167 Medium
167 of 265
Application +4 | -1
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at −6∘C- 6^\circ \mathrm{C} will be (Kf  for  water=1.86  K  kg  mol−1).({K_f}\;\mathrm{for}\;\mathrm{ water} = 1.86\;\mathrm{K}\;\mathrm{ kg }\;\mathrm{mo}{\mathrm{l}^{ - 1}}\mathrm{)}\mathrm{.} and molar mass of ethylene glycol =62  g  mol−1= 62\;g\mathrm{ }\;mo{l^{ - 1}}
View answer and explanation
Correct option: A
804.32  g804.32\mathrm{ }\;g
Option A (804.32  g804.32\mathrm{ }\;g) is correct because it is the mass of ethylene glycol required to lower the freezing point of water by 6°C, calculated using the colligative properties formula.
168 Medium
168 of 265
Understanding +4 | -1
In liquid CCl4{CCl}_4 which of the following has maximum solubility?
View answer and explanation
Correct option: B
Br2{Br}_2
Option B (Br2{Br}_2) is correct because bromine, being a liquid, exhibits greater solubility in liquid CCl4{CCl}_4 compared to solid iodine or gaseous chlorine and sodium chloride is ionic.
169 Medium
169 of 265
Analyzing +4 | -1
Which of the following solution has highest freezing point?
View answer and explanation
Correct option: D
0.1 molal C17H35COONa{C}_{17} {H}_{35} {COONa}
Option D (0.1 molal C17H35COONa{C}_{17} {H}_{35} {COONa}) has the highest freezing point because it has the lowest value of i*molality, resulting in the smallest freezing point depression.
170 Medium
170 of 265
Application +4 | -1
What is the molality of pure water?
View answer and explanation
Correct option: C
55.5
Option C (55.5) is correct because the molality of pure water is calculated by dividing 1000g by the molar mass of water (18 g/mol).
171 Medium
171 of 265
Application +4 | -1
Calculate the osmotic pressure of 0.1NNa3PO40.1 N {Na}_3 {PO}_4 solution at 300 K300 {~K}.
View answer and explanation
Correct option: A
0.821
Option A (0.821) is correct because it is the osmotic pressure calculated using the van't Hoff factor and the osmotic pressure formula.
172 Medium
172 of 265
Application +4 | -1
The vapour pressure of a saturated solution of sparingly soluble salt (XCl3)\left({XCl}_3\right) was 17.20 mm Hg17.20 \mathrm{~mm} \mathrm{~Hg} at 27∘C.27^{\circ} \mathrm{C}. If the vapour pressure of pure water is 17.25 mm Hg17.25 \mathrm{~mm} \mathrm{~Hg} at 27∘C,27^{\circ} \mathrm{C}, then what is the solubility of the sparingly soluble salt XCl3{XCl}_3 in mole per litre?
View answer and explanation
Correct option: A
4.04×10−24.04 \times 10^{-2}
Option A (4.04×10−24.04 \times 10^{-2}) is correct because the solubility is calculated using Raoult's law and the given vapor pressures.
173 Medium
173 of 265
Understanding +4 | -1
When benzoic acid dissolve in benzene, the observed molecular mass is
View answer and explanation
Correct option: A
244
Option A (244) is correct because benzoic acid dimerizes in benzene due to hydrogen bonding, effectively doubling its molecular weight (122 x 2 = 244).
174 Medium
174 of 265
Application +4 | -1
0.48 g0.48 \mathrm{~g} of a substance is dissolved in 10.6 g10.6 \mathrm{~g} of C6H6.\mathrm{C}_6 \mathrm{H}_6. The freezing point of benzene is lowered by 1.8∘C.1.8^{\circ} \mathrm{C}. What will be the molecular weight of the substance (Kf for benzene=5)?\left(K_{\mathrm{f}} \text{ for benzene}=5\right)?
View answer and explanation
Correct option: C
125.79
Option C (125.79) is correct because it is calculated using the depression in freezing point formula, considering the given values for the substance and benzene.
175 Medium
175 of 265
Analyzing +4 | -1
One mole each of urea, glucose and sodium chloride were dissolved in one litre of water Equal osmotic pressure will be produced by solutions of
View answer and explanation
Correct option: B
Urea and glucose
Option B (Urea and glucose) is correct because both urea and glucose are non-electrolytes, so they will produce equal osmotic pressures when dissolved in the same volume of solvent at the same concentration.
176 Medium
176 of 265
Understanding +4 | -1
Statement I: If cells of living organisms are placed in pure water, they swell and burst.

Statement II: Due to osmosis, the movement of water molecules into the cell dilutes the salt content.
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 (If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because cells burst due to water entering them by osmosis, diluting the internal salt concentration.
177 Medium
177 of 265
Application +4 | -1
Air was drawn through a solution containing 40 g40 \mathrm{~g} of solute (non-volatile and non-electrolyte) in 100 g100 \mathrm{~g} of water and then through water. The loss of mass of water was 0.05 g0.05 \mathrm{~g} and the total mass of water absorbed in sulphuric acid tube was 2.05 g.2.05 \mathrm{~g}. The molecular mass of the solute is
View answer and explanation
Correct option: B
288
The correct answer is 288 because the molecular mass of the solute is calculated using Raoult's Law and the given mass loss and absorption data.
178 Medium
178 of 265
Application +4 | -1
24 g24 {~g} of difluoroacetic acid in 400  g400 \,{~g} of water has observed freezing point depression by 1.4  K1.4 \,{~K}. The van't Hoff factor for difluoroacetic acid is
(Kf\left({K}_{{f}}\right. of water =1.86  K  kg mol−1=1.86 \,{~K}\, {~kg} {~mol}^{-1}; Atomic weight of F=19{F}=19 )
View answer and explanation
Correct option: D
1.2
Option D (1.2) is correct because the van't Hoff factor is calculated using the freezing point depression formula, and the result is approximately 1.2.
179 Medium
179 of 265
Understanding +4 | -1
AN aqueous solution is 1.00  \, molal in  KI\,{KI}. Which change will cause the vapour pressure of the solution to increase?
View answer and explanation
Correct option: A
Addition of water
Option A is correct because adding water increases the mole fraction of the solvent, which increases the vapor pressure of the solution according to Raoult's law.
180 Medium
180 of 265
Remembering +4 | -1
What is the composition of gases in the tanks used by the scuba divers?
View answer and explanation
Correct option: B
11.7%11.7 \% helium 56.2%56.2 \% nitrogen 32.1%32.1 \% oxygen
Option B (11.7%11.7 \% helium 56.2%56.2 \% nitrogen 32.1%32.1 \% oxygen) is correct because scuba divers use a mixture of helium, nitrogen, and oxygen with these approximate percentages to avoid nitrogen narcosis at depth.
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