The vapour pressure of pure CCl_4 (molar mass | vedclass

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Question

(A) Given,vapour pressure of pure $CCl_4$ at $25^{\circ} C$ $(p^{\circ}_{CCl_4}) = 115.0 \ torr$. Vapour pressure of pure $SnCl_4$ at $25^{\circ} C$ $

Vedclass · Accepted Answer

$185.85$

Vedclass · Answer

(A) Given,vapour pressure of pure $CCl_4$ at $25^{\circ} C$ $(p^{\circ}_{CCl_4}) = 115.0 \ torr$. Vapour pressure of pure $SnCl_4$ at $25^{\circ} C$ $(p^{\circ}_{SnCl_4}) = 238.0 \ torr$. Moles of $CCl_4$ $(n_{CCl_4}) = \frac{10 \ g}{154 \ g \ mol^{-1}} \approx 0.0649 \ mol$. Moles of $SnCl_4$ $(n_{SnCl_4}) = \frac{15 \ g}{170 \ g \ mol^{-1}} \approx 0.0882 \ mol$. Total moles $(n_{total}) = 0.0649 + 0.0882 = 0.1531 \ mol$. Mole fraction of $CCl_4$ $(\chi_{CCl_4}) = \frac{0.0649}{0.1531} \approx 0.424$. Mole fraction of $SnCl_4$ $(\chi_{SnCl_4}) = 1 - 0.424 = 0.576$. Total vapour pressure $(p_{total}) = \chi_{CCl_4} \cdot p^{\circ}_{CCl_4} + \chi_{SnCl_4} \cdot p^{\circ}_{SnCl_4}$. $p_{total} = (0.424 	imes 115.0) + (0.576 	imes 238.0) = 48.76 + 137.09 = 185.85 \ torr$. Thus,the correct option is $A$.

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