(4)

Given: $n_X=5, n_Y=10$

$x_X=\frac{5}{5+10}=\frac{5}{15}=\frac{1}{3}$

$x_Y=\frac{10}{5+10}=\frac{10}{15}=\frac{2}{3}$

Given: $p_{\text{total}}=70$ $\text{torr}$

$p_{\mathit{X}}^{\mathit{°}}=63$ $\text{torr}, p_{\mathit{Y}}^{\mathit{°}}=78$ $\text{torr}$

$p=p_X+p_Y=\frac{1}{3}\times 63+\frac{2}{3}\times 78=73$ $\text{torr}$

Vapour pressure of solution i.e., 70 torr is less than that predicted by Raoult's law. Hence, it shows negative deviation.

(1)

Mixture of ethanol and acetone shows positive deviation from Raoult’s law.

In pure ethanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds between them. Due to weakening of interactions, the solution shows positive deviation from Raoult’s law.

(4)

For an ideal solution, ${\Delta }_{\text{mix}}H=0 \text{and} {\Delta }_{\text{mix}}V=0$ at constant T and P.

(2)

Maximum boiling azeotropes are formed by those solutions which show negative deviations from Raoult’s law. ${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{HNO}}_3$ mixture shows negative deviations.

(1)

In case of positive deviation from Raoult’s law, A–B interactions are weaker than those between A–A or B–B, i.e., in this case the intermolecular attractive forces between the solute–solvent molecules are weaker than those between the solute–solute and solvent–solvent molecules. This means that in such solutions, molecules of A (or B) will find it easier to escape than in pure state. This will increase the vapour pressure and result in positive deviation.

(4)

For an ideal solution, $\Delta H_{\text{mix}}=0, \Delta V_{\text{mix}}=0,$

Now, $\Delta U_{\text{mix}}=\Delta H_{\text{mix}}-P\Delta V_{\text{mix}}$

$\therefore$     $\Delta U_{\text{mix}}=0$

Also, for an ideal solution, 

          $p_A=x_A{p}_{\mathit{A}}^{\mathit{o}}, p_B=x_B{p}_{\mathit{B}}^{\mathit{o}}$

$\therefore$    $\Delta P=P_{\text{obs}}-P_{\text{calculated by Raoult's law}}=0$

$\Delta G_{\text{mix}}=\Delta H_{\text{mix}}-T\Delta S_{\text{mix}}$

For an ideal solution, $\Delta S_{\text{mix}}\ne 0$

$\therefore$    $\Delta G_{\text{mix}}\ne 0$

(3)

$p_{\text{Benzene}}=x_{\text{Benzene}} p_{\text{Benzene}}^{\circ }$

$p_{\text{Toluene}}=x_{\text{Toluene}} p_{\text{Toluene}}^{\circ }$

For an ideal 1 : 1 molar mixture of benzene and toluene,

$x_{\text{Benzene}}=\frac{1}{2}$ and  $x_{\text{Toluene}}=\frac{1}{2}$

$p_{\text{Benzene}}=\frac{1}{2}{p}_{\text{Benzene}}^{\circ }=\frac{1}{2}\times 12.8$ $\text{kPa}=6.4$ $\text{kPa}$

$p_{\text{Toluene}}=\frac{1}{2}{p}_{\text{Toluene}}^{\circ }=\frac{1}{2}\times 3.85$ $\text{kPa}=1.925$ $\text{kPa}$

Thus, the vapour will contain a high percentage of benzene as the partial vapour pressure of benzene is higher as compared to that of toluene.