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(a) Why is the vapour pressure of a solution of glucose in water lower than that of water ? (b) A `6.90M` solution of `KOH` in water contains `30%` by mass of `KOH`. Calculate the density of the `KOH` solution. [Molar mass of `KOH=56g "mol"^(-1)]` |
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Answer» Correct Answer - (a) When glucose is dissolved in water, the surface molecules consist of both glucose and water, therefore, escaping tendency of molecules of solvent in vapour decreases, therefore, vapour pressure decreases. (b) `M=(% "by massxxdxx10)/("Mol wt.") rArr 6.90=(30xxdxx10)/(56)` `rArr d=(6.90xx56)/(30xx10)=(386.4)/(300)=1.288g cm^(-3)` |
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