Question

The vapour pressure of pure $CCl _{4}$ (molar mass $=154\, g\, mol ^{-1}$ ) and $SnCl _{4}$ (molar mass $=170 \,g\, mol ^{-1}$ ) at $25^{\circ} C$ are $115.0$ and $238.0$ torr respectively. Assuming ideal behaviour, calculate the total approximate vapour pressure in torr of a solution containing $10\, g$ of $CCl _{4}$ and $15 \,g$ of $SnCl _{4}$

• A. 185.85
• B. 190.0
• C. 180.7
• D. 182.1

Answer 1

Answer: (A)

Given, vapour pressure of pure $CCl _{4}$ at $25^{\circ} C\left(p_{ CCl _{4}}^{\circ}\right)=115.$ torr

Vapour pressure of pure $SnCl _{4}$ at $25^{\circ} C\left(p_{ SnCl _{4}}^{\circ}\right)=238.0$ torr

Molar mass of $CCl_{4}\left[M_{\left( CCl _{4}\right)}\right]=154\, g \, mol ^{-1}$

Molar mass of $SnCl_{4}\left[M_{ SnCl _{4}}\right]=170\, g \, mol ^{-1}$

Mass of $ CCl _{4}$ in solution $\left(w_{ CCl _{4}}\right)=10 \, g$

Mass of $SnCl_{4}$ in solution $\left(w_{ SnCl _{4}}\right)=15\, g$

From Dalton's law of partial pressure,

$p_{\text {total }}=\chi_{ CCl _{4}} \cdot p_{ CCl _{4}}^{\circ}+\chi_{ SnCl _{4}} \cdot p_{ SnCl _{4}}^{\circ}$

and $\chi=\frac{n_{\text {simple }}}{n_{\text {total }}}=\frac{w}{M \times n_{\text {total }}}$

Now, $n=\left[\frac{w}{M}\right]_{ CCl _{4}}=\frac{10}{154}=0.065$

and $\left[\frac{w}{M}\right]_{ SnCl _{4}}=\frac{15}{170}=0.09$

Thus, $\chi_{ CCl _{4}}=\frac{n_{ CCl _{4}}}{n_{ CCl _{4}}+n_{ SnCl _{4}}}$

$=\frac{0.065}{0.06+0.09}$

$\chi_{ CCl _{4}}=\frac{0.065}{0.15}=0.43$

Therefore, $\chi_{ SnCl _{4}}=1-0.43=0.57$

Thus, $p_{\text {total }}=0.43 \times 115.00+0.57 \times 238$

$=49.45+135.66$

$=185.11 \approx 185.85$ torr