Question

Calculate the mass of NaCl (molar mass = 58.5 g mol⁻¹) to be dissolved in 37.2 g of water to lower the freezing point by 2°C, assuming that NaCl undergoes complete dissociation. (K_f for water = 1.86 K kg mol⁻¹)

Answer 1

NaCl undergoes complete dissociation as:

NaCl → Na⁺ + Cl⁻

The Van't Hoff factor, i is given as:

\[i = \frac{Number of particles after dissociation}{Number of particles before dissociation}\]
\[ \Rightarrow i = \frac{2}{1} = 2\]

The depression in freezing point of a solution is given by 

\[\Delta T_f = i K_f \frac{w_s \times 1000}{M_s \times W} \ldots . . (i)\]
\[Given: \]
\[ \text{K_f for water} = 1 . 86 K kg {mol}^{- 1} \]
\[\text{Molar mass of solute,} M_s = 58 . 5 g {mol}^{- 1} \]
\[\text{Mass of water,} W = 37 . 2 g\]
\[\Delta T_f = 2^o C = 2 K\]
\[\text{Mass of solute}, w_s = ?\]
\[\text{Substituting the above values in (i), we get}\]
\[2 = 2 \times 1 . 86 \times \frac{w_s \times 1000}{58 . 5 \times 37 . 2}\]
\[ w_s = 1 . 17 g\]
\[\text{Hence, the required mass of NaCl is} 1 . 17 g .\]