Vapour pressure of chloroform (CHCl_3) and di | vedclass

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Question

(D) $1$. Calculate the moles of each component:$n_{CHCl_3} = \frac{25.5 \ g}{119.5 \ g/mol} \approx 0.213 \ mol$$n_{CH_2Cl_2} = \frac{40 \ g}{85 \ g/m

Vedclass · Accepted Answer

None of the above

Vedclass · Answer

(D) $1$. Calculate the moles of each component:$n_{CHCl_3} = \frac{25.5 \ g}{119.5 \ g/mol} \approx 0.213 \ mol$$n_{CH_2Cl_2} = \frac{40 \ g}{85 \ g/mol} \approx 0.471 \ mol$$2$. Calculate the mole fractions:$n_{total} = 0.213 + 0.471 = 0.684 \ mol$$x_{CHCl_3} = \frac{0.213}{0.684} \approx 0.311$$x_{CH_2Cl_2} = \frac{0.471}{0.684} \approx 0.689$$3$. Calculate the total vapour pressure using Raoult's Law:$P_{total} = P^0_{CHCl_3} \cdot x_{CHCl_3} + P^0_{CH_2Cl_2} \cdot x_{CH_2Cl_2}$$P_{total} = (200 \ mm \ Hg 	imes 0.311) + (41.5 \ mm \ Hg 	imes 0.689)$$P_{total} = 62.2 + 28.5935 = 90.7935 \ mm \ Hg$Since $90.7935 \ mm \ Hg$ is not among the options,the correct choice is $D$.

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