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Addition of non-volatile solute to solve...

Addition of non-volatile solute to solvent lowers its vapoure pressure. Therefore, the vapour pressure of a solution (i.e, V.P. of solvent in a solution) is lower than that of pure solvent in a solution) is lower than that of pure solvent, at the same temperature. A higher temperature is needed to raise the vapour pressure upto one atmosphere pressure, when boiling point is attined. However, increase in b.pt. is small . for example, `0.1` molal aqueous sucrose solution boils at `10.05^(@)C`
Sea water, an aqueous solution, which is rich in `Na^(+)` and `Cl^(-)` ions, freezes about `1^(@)C` lower than frozen water . At the freezing point of a pure-solvent, the reates at which two molecule stick together to form the solid and leave it to return to liquid are equal when solute is present. Few solvent molecules are in contact with surface of solid. However, the rate at which the solvent molecules leave, surface of solid remains unchanged. That is why, temperature is lowered to restore the equalibrium. The freezing depression in a dilute solution is proportional to molality of the solute.
When `250 mg` of eugonal is added to `100 g` of camhor `(K_(f) = 29.7 K "molality"^(-1))`. it lowered the freezing point by `0.62^(@)C`. The molar mass of eugonal is :

A

`1.6 xx 10^(2) g//"mol"`

B

`1.6 xx 10^(4) g//"mol"`

C

`1.6 xx 10^(3) g//"mol"`

D

`200 g//"mol"`

Text Solution

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The correct Answer is:
A
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