Embibe Experts Solutions for Chapter: Solutions, Exercise 1: Exercise 1

A solution $\mathrm{M}$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is $0.9$. 

Given,

$\begin{array}{l}{\mathrm{K}}_{\mathrm{f}} \left(\mathrm{water}\right)=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{f}} \left(\mathrm{ethanol}\right)=2.0 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{b}} \left(\mathrm{water}\right)=0.52 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{b}} \left(\mathrm{ethanol}\right)=1.2 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\end{array}$

Standard freezing point of water $=273 \mathrm{K}$

Standard freezing point of ethanol $=155.7 \mathrm{K}$

Standard boiling point of water $=373 \mathrm{K}$

Standard boiling point of ethanol $=351.5 \mathrm{K}$

Vapour pressure of pure water $=32.8 \mathrm{mmHg}$ 

Vapour pressure of pure ethanol $=40 \mathrm{mmHg}$ 

Molecular weight of water $=18 \mathrm{g} {\mathrm{mol}}^{-1}$

Molecular weight of ethanol $=46 \mathrm{g} {\mathrm{mol}}^{-1}$

The vapour pressure of the solution $\mathrm{M}$ is

Two liquids, $\mathrm{X}$ and $\mathrm{Y}$, form an ideal solution. At $300 \mathrm{K}$, the vapour pressure of the solution containing one mole of $\mathrm{X}$ and three moles of $\mathrm{Y}$ is $550 \mathrm{mm} \mathrm{Hg}$ . At the same temperature, if one mole of $\mathrm{Y}$ is further added to this solution, the vapour pressure of the solution increases by $10 \mathrm{mm} \mathrm{Hg}$. The vapour pressure of $\mathrm{X}$ and $\mathrm{Y}$ (in $\mathrm{mm} \mathrm{Hg}$) in their pure states will be, respectively:

The vapour pressure of a solvent decreased by $10 \mathrm{mm} \mathrm{Hg}$ when a nonvolatile solute was added to the solvent. The mole fraction of the solute in the solution is $0.2.$ What would be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \mathrm{mm} \mathrm{Hg}$?

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

$\begin{array}{l}{\mathrm{K}}_{\mathrm{f}}\left(\mathrm{Water}\right)=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{f}}\left(\mathrm{Ethanol}\right)=2.0 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{b}}\left(\mathrm{Water}\right)=0.52 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{b}}\left(\mathrm{Ethanol}\right)=1.2 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\end{array}$

Standard freezing point of water $=273\mathrm{K}$

Standard freezing point of ethanol $=155.7 \mathrm{K}$

Standard boiling point of water $=373 \mathrm{K}$

Standard boiling point of ethanol $=351.5 \mathrm{K}$

Vapour pressure of pure water $=32.8 \mathrm{mmHg}$ 

Vapour pressure of pure ethanol $=40 \mathrm{mmHg}$ 

Molecular weight of water $=18 \mathrm{g} {\mathrm{mol}}^{-1}$

Molecular weight of ethanol$=46 \mathrm{g} {\mathrm{mol}}^{-1}$ 

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]

A solution $\mathrm{M}$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is $0.9$. 

Given,

$\begin{array}{l}{\mathrm{K}}_{\mathrm{f}} \left(\mathrm{water}\right)=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{f}} \left(\mathrm{ethanol}\right)=2.0 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{b}} \left(\mathrm{water}\right)=0.52 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{b}} \left(\mathrm{ethanol}\right)=1.2 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\end{array}$

Standard freezing point of water $=273 \mathrm{K}$

Standard freezing point of ethanol $=155.7 \mathrm{K}$

Standard boiling point of water $=373 \mathrm{K}$

Standard boiling point of ethanol $=351.5 \mathrm{K}$

Vapour pressure of pure water $=32.8 \mathrm{mmHg}$ 

Vapour pressure of pure ethanol $=40 \mathrm{mmHg}$ 

Molecular weight of water $=18 \mathrm{g} {\mathrm{mol}}^{-1}$

Molecular weight of ethanol $=46 \mathrm{g} {\mathrm{mol}}^{-1}$ 

The freezing point of the solution $\mathrm{M}$ is

The $\mathrm{pH}$ of a $0.1 \mathrm{Molar}$ solution of a weak acid $\mathrm{HA}$ is found to be $2$ at a temperature $\mathrm{T}$. The osmotic pressure of the acid solution would be equal to

The degree of dissociation ($\mathrm{\alpha }$) can be calculated using the formula, $\mathrm{\alpha }=\frac{d_{\mathrm{t}}-{\mathrm{d}}_0}{(\mathrm{n}-1){\mathrm{d}}_0}$

This formula is applicable at

$0.004 \mathrm{M} {\mathrm{Na}}_2{\mathrm{SO}}_4$  is isotonic with $0.01 \mathrm{M}$ glucose. Degree of dissociation of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is-