(a) `A` solution is prepared by dissolving `10g` of non-volatile solute in `180g` of `H_(2)O`. If the relative lowering of vapour pressure is `0.005`, find the molar mass of the solute.
(b) How many grams of sucrose `(C_(12)H_(22)O_(11))`must be dissolved in `90g` of water to produce a solution over which the relative humidity is `80%`? Assume the solution is ideal.

Answer

(a) `199 g//"mole"`, (b) `427.5gm`

Explanation

(a) Relative lowering of vapour pressure `=X_(A)`
or `0.005=(10//M_(A))/(10//M_(A)+180//18)` (Let `M_(A)` is the molar mass of solute)
so `M_(A)=199 gm//"mole"`
(b) Let `P^(@)` be the vapour pressure of water, now if relative humidity is `80%` after dissolving `C_(12)H_(22)O_(11)`. then the new pressure of solution will be `P` (let).
so `(P)/(P_(0))=0.8` or `P=0.8P_(0)`
Now Relative lowring of pressure `=(P_(0)-P)/(P_(0))=X_(A)`
so `(P_(0)-0.8P_(0))/(P_(0))=(x)/(342/(x/(342)+(90)/(18)))` {let `x` be the `wt.` of `C_(12)H_(22)O_(11)` dissolved}
`therefore x=427.5gm`.