Embibe Experts Solutions for Chapter: Solutions, Exercise 3: Level 3

Author:Embibe Experts

Embibe Experts Chemistry Solutions for Exercise - Embibe Experts Solutions for Chapter: Solutions, Exercise 3: Level 3

Attempt the practice questions on Chapter 22: Solutions, Exercise 3: Level 3 with hints and solutions to strengthen your understanding. Chemistry Crash Course JEE Main solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Solutions, Exercise 3: Level 3 with Hints & Solutions

MEDIUM
JEE Main
IMPORTANT

The relative lowering of vapour pressure of an aqueous solution of urea is 0.018. If Kb for H2O is 0.54 °C m-1, the elevation in boiling point will be:

HARD
JEE Main
IMPORTANT

Insulin C2H10O5n is dissolved in a suitable solvent and the osmotic pressure (π) of solutions of various concentrations Cin g/cm3 is measured at 20°C. The slope of a plot of π against C is found to be 4.65×10-3. The molar mass of the insulin is

HARD
JEE Main
IMPORTANT

At 300 K, vapour pressure of substance A is 0.95 atm and vapour pressure of substance B is 0.15 atm. A solution of A and B is prepared and allowed to equilibrate with its vapour. The vapour is found to have equal moles of A and B. What is lhe mole fraction of A in the original solution?

MEDIUM
JEE Main
IMPORTANT

Which of the following relations is correct for the mole fraction of the solute (x) ? (m and M  are the molality of the solution and molar mass of the solvent respectively.)

HARD
JEE Main
IMPORTANT

A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 . Given

Kf(Water)=1.86 K kg mol-1Kf(Ethanol)=2.0 K kg mol-1Kb(Water)=0.52 K kg mol-1Kb(Ethanol)=1.2 K kg mol-1

Standard freezing point of water =273K

Standard freezing point of ethanol =155.7 K

Standard boiling point of water =373 K

Standard boiling point of ethanol =351.5 K

Vapour pressure of pure water =32.8 mmHg 

Vapour pressure of pure ethanol =40 mmHg 

Molecular weight of water =18 g mol-1

Molecular weight of ethanol=46 g mol-1 

Water is added to the solution M such that the mole fraction of water in solution becomes 0.9. What is the boiling point of this solution?

[Hint: Solute is ethanol and solvent is water]

HARD
JEE Main
IMPORTANT

A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9

Given,

Kf water=1.86 K kg mol-1Kf ethanol=2.0 K kg mol-1Kb water=0.52 K kg mol-1Kb ethanol=1.2 K kg mol-1

Standard freezing point of water =273 K

Standard freezing point of ethanol =155.7 K

Standard boiling point of water =373 K

Standard boiling point of ethanol =351.5 K

Vapour pressure of pure water =32.8 mmHg 

Vapour pressure of pure ethanol =40 mmHg 

Molecular weight of water =18 g mol-1

Molecular weight of ethanol =46 g mol-1

The vapour pressure of the solution M is

HARD
JEE Main
IMPORTANT

A liquid mixture of A and B is placed in a cylinder-and-piston arrangement. The piston is slowly pulled out isothermally, so that the volume of the liquid decreases and that of the vapour increases. At the instant when the quantity of the liquid still remaining is negligibly small, the mole fraction of A in the vapour phase is 0.4 pA0=0.4 atm, pB0=1.2 atm at the temperature in question. Calculate the total pressure at which the liquid has almost evaporated. Assume ideal behaviour.

Give answer after multiplying with 100 and rounding off to the nearest integer value.

MEDIUM
JEE Main
IMPORTANT

A solution of a nonvolatile solute in water freezes at -0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and Kf for water is 1.86 degree/molal. Calculate the magnitude of change in vapour pressure of this solution in mmHg at 298 K.

Give answer after multiplying with 100 and rounding off to the nearest integer value.