Question:
Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol−1.
Solution:
Mass of acetic acid, w1 = 75 g
Molar mass of ascorbic acid (C6H8O6), M2 = 6 × 12 + 8 × 1 + 6 × 16
= 176 g mol−1
Lowering of melting point, ΔTf = 1.5 K
We know that:
$\Delta T_{f}=\frac{K_{f} \times w_{2} \times 1000}{M_{2} \times w_{1}}$
$\Rightarrow w_{2}=\frac{\Delta T_{f} \times M_{2} \times w_{1}}{K_{f} \times 1000}$
$=\frac{1.5 \times 176 \times 75}{3.9 \times 1000}$
= 5.08 g (approx)
Hence, 5.08 g of ascorbic acid is needed to be dissolved.
Note: There is a slight variation in this answer and the one given in the NCERT textbook.