Question

What mass of $NaC$l must be dissolved in 65.0 g of water to lower the freezing of water by ${7.5}^{o}C$ ? The freezing point depression constant ($K_f$) for water 1.860 C/m. Assume Van't Hoff factor for $NaCl$ is 1.87. (Molar mass of $NaCl=$ 58. 5)

Answer 1

$\Delta T_f=i{K}_{f}m$

Here, $\Delta T_f=$ lowering in freezing point of water $=7.500C$

$i=$ Van't Hoff factor $=1.87$

$K_f=1.860C/m$

$m=\text{molality}$

Putting the values of known variables to find $m$,

$m=\frac{\Delta T_f}{i{K}_f}=\frac{7.5}{1.87\times 1.86}=2.156$

$m=$ molality $=\frac{\text{number of moles of NaCl}}{\text{mass of water in kg}}$

$\text{number of moles of NaCl}=\frac{\text{mass of NaCl}}{\text{molar mass of NaCl}}=\frac{\text{mass of NaCl}}{58.5}$

$m=\frac{\text{mass of NaCl}/58.5}{0.065}$

$\text{mass of NaCl}=m\times 0.065\times 58.5=2.156\times 0.065\times 58.5$

$\text{mass of NaCl}=8.199g$