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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. `141.93mm`B. `94.39mm`C. `199.34mm`D. `143.99mm` |
Answer» Correct Answer - A `W_(B) = 0.5g` `M_(B) = 65` amu `W_(A) = "vol" xx "density" = 100 xx 1.58 = 158g` `M_(A)(C CI_(4)) = 12 +35.5 xx 4 = 154` `(P^(@)-P_(s))/(P^(@)) = (W_(B))/(M_(B)).(M_(A))/(W_(A))` `P^(@) - P_(S) = (W_(B)M_(A))/(M_(B)W_(A)) xx P^(@)` `P_(S) = P^(@) - (W_(B)M_(A))/(M_(B)W_(A)) xx P^(@)` `= 143 - (0.5 xx 154)/(65 xx 158) xx 143` `= 143 - 1.03 = 141.97mm` |
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