Question

Depression in freezing point is $6 \,K$ for $ NaCl $ solution. If $ K_{f} $ for water is $1.86\, K/kg\, mol$, amount of $ NaCl $ dissolved in $1 \,kg$ water is:

• A. 3.42
• B. 1.62
• C. 3.24
• D. 1.71

Answer 1

Answer: (B)

$\Delta T_{f}=i \times K_{f} \times \frac{n}{W} \times 1000$
$ i=$ vant Hoff factor
$K_{f}=1.86\, K / kg \,mol $
$n =$ number of moles
$W =$ weight of solvent $=1\, kg =1000\, g$
$ \Delta T_{f}=6 \,K$ so,
$6=2 \times 1.86 \times \frac{n}{1000} \times 10006$
$=2=1.86 \times n$
$n=\frac{6}{1.86}=1.62$