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The vapour pressure of acetone at `20^(@)C` is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at `20^(@)C`, its vapour pressure was 183 torr. The molar mass of the substance is :A. 128B. 488C. 32D. 64 |
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Answer» Correct Answer - d `(P_(A)^(@)P_(S))/P_(S)=(W_(B)xxW_(A))/(W_(A)xxM_(B))` `((185-183))/((183))=((1.2 g)xx(58 g mol^(-1)))/((100g)xxM_(B))` `M_(B)=((1.2g)xx(58g mol^(-1))xx(183))/(2xx(100g))=63.68g mol^(-1)` =64 g `mol^(-1)` |
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