The vapour pressure of pure benzene...
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The vapour pressure of pure benzene at a certain temperature is 0.950 bar. Anon-volatile, non-electrolyte solid weighing 0.5gm is added to 39.0 gm of benzene, vapour pressure of the solution becomes 0.945 bar. What is the molar mass of the solid substance? (A) 160 gm mol-? (B) 210 gm mol (C) 95 gm mol-? (D) 189 gm mol-

JEE/Engineering Exams
Chemistry
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( p^{0}=0.950 mathrm{bar} quad mathrm{Ps}=0.945 mathrm{bar} ) ( omega=0.5 mathrm{g} quad M_{B}=? ) weight of solvent (ben zene) ( =39.0 mathrm{gm} ) molecular weight of benzere: ( 78 mathrm{gm} ). we know ( frac{p^{0}-p_{s}}{p_{0}}=x_{B}=frac{n_{B}}{n_{A}} ) ( frac{0.950-0.945}{0.950}=frac{48 / m_{3}}{w_{A} / m_{A}}=frac{frac{0.5}{m}}{39 / 782} ) ( m=frac{0.950}{0.005}=190 mathrm{gm} frac{1}{5} mathrm{mol}^{-1} ) Ans D ( [189 mathrm{gm} / mathrm{mol} )
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