Embibe Experts Solutions for Chapter: Solutions, Exercise 2: Level 2

The addition of $0.643 \mathrm{g}$ of a compound to $50 \mathrm{ml}$ of benzene (density $=0.879 \mathrm{g} {\mathrm{mol}}^{-1}$) lowers the freezing point from $5.51°\mathrm{C}$ to $5.03°\mathrm{C}$. If the freezing point constant, ${\mathrm{K}}_{\mathrm{f}}$, for benzene is $5.12 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}$, the molar mass of the compound is approximately:

A solution containing $8.0 \mathrm{g}$ of nicotine in $92 \mathrm{g}$ of water freezes $0.925$ degrees below the normal freezing point of water. If the molal freezing point depression constant, ${\mathrm{K}}_{\mathrm{f}}=1.85°\mathrm{C} {\mathrm{mol}}^{-1}$, then the molar mass of nicotine is:

The vapour pressure of pure liquid $\mathrm{A}$ is $10 \mathrm{torr}$ and at the same temperature when $1 \mathrm{g}$ of $\mathrm{B}$ solid is dissolved in $20 \mathrm{g}$ of $\mathrm{A}$, its vapour pressure is reduced to $9.0 \mathrm{torr}$. If the molecular mass of $\mathrm{A}$ is $200 \mathrm{amu}$, then the molecular mass of $\mathrm{B}$ is:

The vapour pressure of a solution of a non-volatile solute $\mathrm{B}$ in a solvent $\mathrm{A}$ is $95\%$ of the vapour pressure of the solvent at the same temperature. If the molecular weight of the solvent is $0.3$ times the molecular weight of the solute, what is the ratio of weight of solvent to solute?

In $108 \mathrm{g}$ of water, $18 \mathrm{g}$ of a non-volatile compound is dissolved. At $100°\mathrm{C}$ the vapour pressure of the solution is $750 \mathrm{mmHg}$. Assuming that the compound does not undergo association or dissociation, the molar mass of the compound in $\mathrm{g} {\mathrm{mol}}^{-1}$ is

A mixture of toluene and benzene boils at $100°\mathrm{C}$. Assuming ideal behaviour, the mole fraction of toluene in the mixture is closest to:

[Vapour pressures of pure toluene and pure benzene at $100°\mathrm{C}$ are $0.742$ and $1.800 \mathrm{bar}$, respectively. $1 \mathrm{atm}=1.013 \mathrm{bar}$]

$\mathrm{pH}$ of a $0.1 \mathrm{M}$ monobasic acid is found to be $2$. Hence, its osmotic pressure at a given temperature $\mathrm{TK}$ is-

When $20 \mathrm{g}$ of naphthenic acid $\left({\mathrm{C}}_{11}{\mathrm{H}}_8{\mathrm{O}}_2\right)$ is dissolved in $50 \mathrm{g}$ of benzene $\left({\mathrm{K}}_{\mathrm{f}}=1.72 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\right)$, a freezing point depression of $2 \mathrm{K}$ is observed. The van't Hoff factor $\left(\mathrm{i}\right)$ is x, then the value of 10x is