Question

The freezing point of a solution, prepared from $1.25 \,gm$. of a non-electrolyte and $20 \,gm$. of water, is $271.9\, K$. If the molar depression constant is $1.86\, K / mol$, then molar mass of the solute is

• A. 103.3
• B. 105.7
• C. 107.9
• D. 109.3

Answer 1

Answer: (B)

The molar mass of solute $=\frac{1000 \times K_{f} \times W}{\Delta T_{f} \times w}$
Here weight of solute $(W)=1.25\, g$
weight of solvent $(w)=20\, g$
Molar depression constant
$\left(K_{f}\right)=1.86$
Freezing point of a solution
$\left(\Delta T_{f}\right) =(273-271.9) $
$=1.1\, K$
Putting the values, the molar mass of the solute
$=\frac{1000 \times 1.86 \times 1.25}{1.1 \times 20}=105.7$