Vapour pressure of CCl_{4} at 25,^{circ} C is | vedclass

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Question

(A) The relative lowering of vapour pressure is given by the formula: $\frac{p^{\circ} - p_{s}}{p^{\circ}} = x_{2} = \frac{n_{2}}{n_{1} + n_{2}} \appr

Vedclass · Accepted Answer

$141.93\, mm\, Hg$

Vedclass · Answer

(A) The relative lowering of vapour pressure is given by the formula: $\frac{p^{\circ} - p_{s}}{p^{\circ}} = x_{2} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}}$.First,calculate the moles of solute $(n_{2})$: $n_{2} = \frac{0.5}{65} \approx 0.00769 \, mol$.Next,calculate the mass of $CCl_{4}$: $Mass = Density 	imes Volume = 1.58 \, g/cm^{3} 	imes 100 \, cm^{3} = 158 \, g$.Molar mass of $CCl_{4} = 12 + 4 	imes 35.5 = 154 \, g/mol$.Moles of solvent $(n_{1})$: $n_{1} = \frac{158}{154} \approx 1.026 \, mol$.Using the formula: $\frac{143 - p_{s}}{143} = \frac{0.00769}{1.026}$.$143 - p_{s} = 143 	imes 0.007495 = 1.0718$.$p_{s} = 143 - 1.0718 = 141.9282 \, mm\, Hg \approx 141.93 \, mm\, Hg$.

Vapour pressure of $CCl_{4}$ at $25\,^{\circ} C$ is $143\, mm\, Hg$. $0.5\, g$ of a non-volatile solute (mol. wt. $65$) is dissolved in $100\, mL$ of $CCl_{4}$. Find the vapour pressure of the solution. (Density of $CCl_{4} = 1.58\, g / cm^{3}$)

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