Chapter 12, Problem 12.91SP

(a)

Interpretation Introduction

To determine:

Whether $0.\text{5 m}$ sucrose or 0.35 m ${\text{HNO}}_{\text{3}}$ have higher freezing point

EXPLAINATION:

Among 0.50 m sucrose and 0.35 m HNO₃the freezing point depression in sucrose solution will be more than HNO₃solution because of higher molal value.
So HNO₃solution will have higher freezing point.

(b)

Interpretation Introduction

To determine:

Whether $0.\text{5 m}$ sucrose or 0.35 m ${\text{HNO}}_{\text{3}}$ have higher boiling point

EXPLAINATION:

While we compare 0.50 m sucrose and 0.35 m HNO₃in terms of higher boiling Point, sucrose solution will be the one with higher boiling point, as Elevation in boiling point upon solute addition to pure solvent depends upon the molal value of solution. More the molality of solution more will be the elevation and hence more will be the boiling point of sth solution.

  $\Delta {\text{T}}_{\text{B}}={\text{ K}}_{\text{B}}\times \text{m}$

Sucrose solution will have more boiling point because of high molality value.

(c)

Interpretation Introduction

To determine:

Whether $0.\text{5 m}$ sucrose or 0.35 m ${\text{HNO}}_{\text{3}}$ have lower vapor pressure.

EXPLAINATION:

By Raoult’s Law Vapour pressure the P?_solventvalue will be same so the solution vapour pressure will depend Solely upon mole fraction of solvent i.e (??_solvent)
.
Mole fraction of surose solution = $0.\text{5}/\left(0.\text{5}+\text{55}.\text{6}\right)$

Where 55.6 is moles of water in solution.
Mole fraction of HNO₃ solution = $0.\text{35}/\left(0.\text{35}+\text{55}.\text{6}\right)$

As we see numerator value for HNO₃solution is lesser than the other as a result of which the vapour pressure for nitric acid solution will be lower than HNO₃
.