A solution is prepared by mixing 8.5 g of CH | vedclass

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(C) Molar mass of $CHCl_3 = 12 + 1 + 3(35.5) = 119.5 \ g \ mol^{-1}$.Molar mass of $CH_2Cl_2 = 12 + 2 + 2(35.5) = 85 \ g \ mol^{-1}$.Moles of $CHCl_3

Vedclass · Accepted Answer

$0.325$

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(C) Molar mass of $CHCl_3 = 12 + 1 + 3(35.5) = 119.5 \ g \ mol^{-1}$.Molar mass of $CH_2Cl_2 = 12 + 2 + 2(35.5) = 85 \ g \ mol^{-1}$.Moles of $CHCl_3 = \frac{11.95 \ g}{119.5 \ g \ mol^{-1}} = 0.1 \ mol$.Moles of $CH_2Cl_2 = \frac{8.5 \ g}{85 \ g \ mol^{-1}} = 0.1 \ mol$.Total moles $= 0.1 + 0.1 = 0.2 \ mol$.Mole fraction of $CHCl_3$ $(x_{CHCl_3}) = \frac{0.1}{0.2} = 0.5$.Mole fraction of $CH_2Cl_2$ $(x_{CH_2Cl_2}) = \frac{0.1}{0.2} = 0.5$.Partial pressure of $CHCl_3$ $(P_{CHCl_3}) = x_{CHCl_3} 	imes P^{\circ}_{CHCl_3} = 0.5 	imes 200 = 100 \ mm \ Hg$.Partial pressure of $CH_2Cl_2$ $(P_{CH_2Cl_2}) = x_{CH_2Cl_2} 	imes P^{\circ}_{CH_2Cl_2} = 0.5 	imes 415 = 207.5 \ mm \ Hg$.Total pressure $(P_{total}) = P_{CHCl_3} + P_{CH_2Cl_2} = 100 + 207.5 = 307.5 \ mm \ Hg$.Mole fraction of $CHCl_3$ in vapour phase $(y_{CHCl_3}) = \frac{P_{CHCl_3}}{P_{total}} = \frac{100}{307.5} \approx 0.325$.

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