Question

$6g$ of a non volatile, non electrolyte X dissolved in 100g of water freezes at $-{0.93}^{0}C$. The molar mass of X in $gmo{l}^{-1}$ is:

$(K_{f}of{H}_{2}O=1.86KKgmo{l}^{-1})$

• A. $60$
• B. $140$
• C. $180$
• D. $120$

Answer 1

The correct option is D $120$

Freezing point is the temperature at which liquid and solid remain in equilibrium and have equal vapour pressure.

$\Delta T_f(DepressioninF.P)=F.Pofsolvent-F.Pofsolution----\left(1\right)$

$\Delta T_f=K_f\times m----\left(2\right)$

Where,

m = molality

$K_f$ = Cryoscopic constant

from equation (1) and (2)

$T_f^{\circ }-T_{f\left(sol\right)}=K_f\times m$

$00-(-0.93)=1.86\times \frac{\left(massofx\right)\times 1000}{\left(Molarmassofx\right)\times \left(Massof{H}_{2}0\right)}$

$0.93=1.86\times \frac{6\times 1000}{\left(Molarmassofx\right)\times 100}$

$Molarmassofx=2\times 6\times 10$

$Molarmassofx=120gmo{l}^{(-1)}$