Embibe Experts Solutions for Chapter: Solutions, Exercise 4: EXERCISE-4

$30 \mathrm{mL}$ of ${\mathrm{CH}}_3\mathrm{OH}\left(\mathrm{d}=0.7980 \mathrm{g}{ \mathrm{cm}}^{-3}\right)$ and $70 \mathrm{mL}$ of ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{d}=0.9984 \mathrm{g}{ \mathrm{cm}}^{-3}\right)$ are mixed at $25°\mathrm{C}$ to form a solution of density $0.9575 \mathrm{g}{ \mathrm{cm}}^{-3}$. Calculate the freezing point of the solution. ${\mathrm{K}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)$ is $1.86 \mathrm{kg}{ \mathrm{mol}}^{-1} \mathrm{K}$. Also calculate its molarity.

The freezing point depression of a $0.109 \mathrm{M}$ aq. solution of formic acid is $-0.21°\mathrm{C}$. Calculate the equilibrium constant for the reaction,
$\mathrm{HCOOH}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{HCOO}}^{\ominus }\left(\mathrm{aq}\right)$
${\mathrm{K}}_{\mathrm{f}}$ for water $=1.86 \mathrm{kg}{ \mathrm{mol}}^{-1} \mathrm{K}$

The freezing point of $0.02 \mathrm{mol}$ fraction solution of acetic acid $\left(\mathrm{A}\right)$ in benzene $\left(\mathrm{B}\right)$ is $277.4 \mathrm{K}$. Acetic acid exists partly as a dimer $2 \mathrm{A}={\mathrm{A}}_2$. Calculate equilibrium constant for the dimerisation. Freezing point of benzene is $278.4 \mathrm{K}$ and its heat of fusion ${\mathrm{\Delta H}}_{\mathrm{f}}$ is $10.042 \mathrm{kJ}{ \mathrm{mol}}^{-1}$.

Vapour pressure of ${\mathrm{C}}_6{\mathrm{H}}_6$ and ${\mathrm{C}}_7{\mathrm{H}}_8$ mixture at $50°\mathrm{C}$ is given by $\mathrm{P}\left(\mathrm{mmHg}\right)=179{\mathrm{X}}_{\mathrm{B}}+92$, where ${\mathrm{X}}_{\mathrm{B}}$ is the mole fraction of ${\mathrm{C}}_6{\mathrm{H}}_6$. A solution is prepared by mixing $936 \mathrm{g}$ benzene and $736 \mathrm{g}$ toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of $50°\mathrm{C}$, what would be mole fraction of ${\mathrm{C}}_6{\mathrm{H}}_6$ in the vapour state?

The vapour pressure of a certain liquid is given by the equation :
${\log }_{10}\mathrm{P}=3.54595-\frac{313.7}{\mathrm{T}}+1.40655{\log }_{10}\mathrm{T}$ where $\mathrm{P}$ is the vapour pressure in $\mathrm{mm}$ and $\mathrm{T}=\text{Kelvin}$
Temperature. Determine the molar latent heat of vaporisation as a function of temperature. Calculate the its value at $80 \mathrm{K}$.

A very dilute saturated solution of a sparingly soluble sali ${\mathrm{A}}_3{\mathrm{B}}_4$ has a vapour pressure of $20 \mathrm{mm}$ of $\mathrm{Hg}$ at temperature $\mathrm{T}$, while pure water exerts a pressure of $20.0126 \mathrm{mm} \mathrm{Hg}$ at the same temperature. Calculate the solubility product constant of ${\mathrm{A}}_3{\mathrm{B}}_4$ at the same temperature.

An ideal solution was prepared by dissolving some amount of cane sugar (non-volatile) in $0.9$ moles of water. The solution was then cooled just below its freezing temperature $\left(271 \mathrm{K}\right)$, where some ice get separated out. The remaining aqueous solution registered a vapour pressure of 700 torr at $373 \mathrm{K}$. Calculate the mass of ice separated out, if the molar heat of fusion of water is $96 \mathrm{kJ}$.

Tritium, $\mathrm{T}$ (an isotope of $\mathrm{H}$) combines with fluorine to form weak acid $\mathrm{TF}$, which ionizes to give ${\mathrm{T}}^{+}$. Tritium is radioactive and is a $\mathrm{\beta }$-emitter. A freshly prepared aqueous solution of $\mathrm{TF}$ has $\mathrm{pT}$ (equivalent of $\mathrm{pH}$) of $1.5$ and freezes at $-0.372°\mathrm{C}$. If $600 \mathrm{mL}$ of freshly prepared solution were allowed to stand for $24.8$ years, calculate (i) ionization constant of $\mathrm{TF}$. (ii) Number of $\mathrm{\beta }$-particles emitted.
(Given ${\mathrm{K}}_{\mathrm{f}}$ for water $=1.86 \mathrm{kg} \mathrm{mol}{ \mathrm{K}}^{-1},{\mathrm{t}}_{1/2}$ for tritium $=12.4$ years.)