Vapour pressure of CCl_4 at 25 text{ }^circte | vedclass

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Question

(A) $1$. Calculate mass of $CCl_4$: $	ext{Mass} = 	ext{Density} 	imes 	ext{Volume} = 1.58 	ext{ g/cm}^3 	imes 100 	ext{ cm}^3 = 158 	ext{ g}$.

Vedclass · Accepted Answer

$141.93$

Vedclass · Answer

(A) $1$. Calculate mass of $CCl_4$: $	ext{Mass} = 	ext{Density} 	imes 	ext{Volume} = 1.58 	ext{ g/cm}^3 	imes 100 	ext{ cm}^3 = 158 	ext{ g}$.$2$. Calculate moles of $CCl_4$ $(M_{CCl_4} = 12 + 4 	imes 35.5 = 154 	ext{ g/mol})$: $n_{CCl_4} = \frac{158}{154} \approx 1.026 	ext{ mol}$.$3$. Calculate moles of solute: $n_{	ext{solute}} = \frac{0.5}{65} \approx 0.00769 	ext{ mol}$.$4$. Calculate mole fraction of solvent $(X_A)$: $X_A = \frac{n_{CCl_4}}{n_{CCl_4} + n_{	ext{solute}}} = \frac{1.026}{1.026 + 0.00769} = \frac{1.026}{1.03369} \approx 0.99256$.$5$. Calculate vapour pressure of solution $(P_s)$: $P_s = P^0 	imes X_A = 143 	ext{ mm Hg} 	imes 0.99256 \approx 141.93 	ext{ mm Hg}$.

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