The vapour pressure of a 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 up to the atmospheric pressure, when boiling begins. However, increase is small, like `0.1 "mol"kg^(-1)` aqueous sucrose solution boils at `100.05^(@)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 rates at which two molecules stick together to form the solid and leave it to return to liquid state are equal when solute is present. Fewer solvent molecules are in contact 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 freezing point depression in an ideal solution is proportional to molality of the solute.
Whene 250 m,g of eugenol is added to 100 gm of camphor `(K_(f)=37.9)`, it lowered the freezing point by `0.62^(@)C`, the molar mass of eugenol is

Options

A. `1.6xx10^(2)`gm/mole
B. `1.6xx10^(4)` gm/mole
C. `1.6xx10^(3)` gm/mole
D. 200 gm/mole

Explanation

To find the molar mass of eugenol when added to camphor, we will use the formula for freezing point depression, which is given by:

\[ \Delta T_f = K_f \cdot m \]

where:
- \(\Delta T_f\) is the depression in freezing point,
...