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InterviewSolution
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The vapour pressure of pure benzene at a certain temperature is `0.850` bar. A non-volatile, non-electrolyte solid weighting `0.5 g` when added to `39.0 g` of benzene (molar mass `78 g mol^(-1)`). The vapour pressure of the solution then is `0.845` bar. What is the molar mass of the solid substance?A. 170 amuB. 160 amuC. 120 amuD. 190 amu |
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Answer» Correct Answer - 1 we are given `P_(1)^(0)=0.850` bar, `P_(1)`=0.845 bar `M_(1)=78 g mol^(-1), W_(1)=39 g and W_(2)=0.5 g` Substituting these values in Equation `(2.58)`, we get `DeltaP_(1)//P_(1)^(0)=(W_(2)xxM_(1))/(W_(1)xxM_(2))` `((0.850 - 0.845) "bar")(0.850 "bar")=(0.5 gxx78 g mol^(-1))/(M_(2)xx39 g)` or `M_(2)=170 g mol^(-1)` Thus, molecular mass of noncolatile solid is 170 amu |
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