Question

The amount of urea to be dissolved in $500cc$ of water $(K_f={1.86}^{\circ }Ckgmo{l}^{-1})$ to produce a depression of ${0.186}^{\circ }C$ in the freezing point is:
( Molar mass of urea = $60g/mol$)

• A. $9$
• B. $6g$
• C. $3g$
• D. $0.5g$

Answer 1

The correct option is C $3g$

Freezing point depression of a solution is given by,
$△T_f=K_f\times m$
where,
$△T_f$ is the depression in freezing point.
$K_f$ is molal depression constant.
m is molality of the solution.
From the above equation,
We know that,
$\Delta T=K_f\times \frac{w\times 1000}{M\times W}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute
Given:
$1cc\approx mL$
Density of water is $1g/mL$
Thus,
$\text{Weight of water}=500\times 1\Rightarrow 500g$
$\text{Molar mass of urea,}M=60gmo{l}^{-1}$
$\text{Depression in freezing point,}△T_f={0.186}^{\circ }C$
$\text{Molal depression constant,}{K}_f={1.86}^{\circ }Ckgmo{l}^{-1}$
Substituting the given values in above equation gives,
$0.186=1.86\times \frac{w\times 1000}{60\times 500}w=3g$
Thus, the amount of urea dissolved is $3g$