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When we add a non-volatile solute to a solvent lowers its vapour pressure. Therefore, the vapour pressure of 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 attained. However, b.pt. increases small. For example, 0.1 molal aqueous sucrose solution boils at `100.05^(@)C`. Sea water, which is rich in `Na^(+)` and `Cl^(-)` ions, freezes about `1^(@)C` lower than freezes pure water. At the freezing point of a pure-solvent, the rates at which two molecule stick together to form the solute is present. Few solvent molecules are in constant with surface of solid. However, the rate at which the solvent molecules leave the surface of solid remains unchanged. That is why temperature is lowered to restore the equilibrium. The depression is lowered to restore the equilibrium. The depression in freezing point in a dilute solution is proportional to molality of the solute. An aqueous solution of 0.1 molal concentration of sucrose should have freezing point `(K_(f)=1.86 K mol^(-1)kg)`A. `0.186^(@)C`B. `1.86^(@)C`C. `-1.86^(@)C`D. `-0.186^(@)C`

Answer» Correct Answer - D
`Delta T_(f)=K_(f)m=1.86xx0.1=0.186`, Freezing point
`=0-Delta T_(f)=-0.186^(@)C`


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