Question

The vapour pressure of pure $CC{l}_4$ (molar mass $=154gmo{l}^{-1}$ ) and $SnC{l}_4$ (molar mass $=170gmo{l}^{-1}$ ) 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 $10g$ of $CC{l}_4$ and $15g$ of $SnC{l}_4$

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

Answer 1

Answer: (A)

Given, vapour pressure of pure $CC{l}_4$ at $25^{\circ }C\left(p_{CC{l}_4}^{\circ }\right)=115.$ torr

Vapour pressure of pure $SnC{l}_4$ at $25^{\circ }C\left(p_{SnC{l}_4}^{\circ }\right)=238.0$ torr

Molar mass of $CC{l}_4\left[M_{\left(CC{l}_4\right)}\right]=154gmo{l}^{-1}$

Molar mass of $SnC{l}_4\left[M_{SnC{l}_4}\right]=170gmo{l}^{-1}$

Mass of $CC{l}_4$ in solution $\left(w_{CC{l}_4}\right)=10g$

Mass of $SnC{l}_4$ in solution $\left(w_{SnC{l}_4}\right)=15g$

From Dalton's law of partial pressure,

$p_{\text{total }}={\chi }_{CC{l}_4}\cdot p_{CC{l}_4}^{\circ }+{\chi }_{SnC{l}_4}\cdot p_{SnC{l}_4}^{\circ }$

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

Now, $n={\left[\frac{w}{M}\right]}_{CC{l}_4}=\frac{10}{154}=0.065$

and ${\left[\frac{w}{M}\right]}_{SnC{l}_4}=\frac{15}{170}=0.09$

Thus, ${\chi }_{CC{l}_4}=\frac{n_{CC{l}_4}}{n_{CC{l}_4}+n_{SnC{l}_4}}$

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

${\chi }_{CC{l}_4}=\frac{0.065}{0.15}=0.43$

Therefore, ${\chi }_{SnC{l}_4}=1-0.43=0.57$

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

$=49.45+135.66$

$=185.11\approx 185.85$ torr