Question

How many grams of HCl must be added to 500 mL water to produce a solution that freezes at $-{1.86}^{o}C$? (molal freezing constant = ${1.86}^{o}Ckg/mol$).

• A. $4.6$
• B. $9.1$
• C. $18.3$
• D. $36.5$
• E. $73.0$

Answer 1

The correct option is B $9.1$

Let $W$ grams of $HCl$ must be added to 500 mL water. The molecular weight of $HCl$ is 36.5 g/mol.

The number of moles of $HCl$ $n=\frac{W}{36.5}$ mol.

The mass of water

$=500mL\times 1g/mL=500g=500g\times \frac{1kg}{1000g}=0.500kg$
The molality of $HCl$ $m=\frac{W}{36.5\times 0.500}=\frac{W}{18.25}m$

The freezing point of pure water is 0 deg C. The solution freezes at $-{1.86}^{o}C$.

The depression in the freezing point $\Delta T_f=0^{o}C-(-{1.86}^{o}C)={1.86}^{o}C$.

The depression in the freezing point
$\Delta T_f=i\times Kf\times m$
${1.86}^{o}C=2\times {1.86}^{o}Ckg/mol\times \frac{W}{18.25}m$

$W=9.1g$

Note: The vant Hoff factor 'i' for $HCl$ is 2 as one $HCl$ molecule completely ionises to provide 2 ions.