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Vapour pressure of `"CCl"_(4)` at `25^(@)C` is 143 mm Hg . 0.5 g of a non-volatile solute ( molar mass = `65 mol^(-1)`) is dissolved in 100 mL of `"CCl"_(4)` (density = 1.538g `mL^(-1)` ) Vapour pressure of solution is :A. 141.93 mmB. 94.39 mmC. 199.34 mmD. 143.99 mm |
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Answer» Correct Answer - A `(P_(1)^(o)-P)/P_(1)^(o)=n_(2)/n_(1)` `(143-P)/143=(0.5//65)/(158//154)` or P=141.93 mm |
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