Ncert Exercises & Solutions

2.11 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. K_f = 3.9 K kg mol⁻¹.

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, ∆T_f = 1.5 K

We know that:

$∆ T_{f} = \frac{K_{f} \times w_{2} \times 1000}{M_{2} \times w_{1}}$
$\Rightarrow w_{2} = \frac{∆ T_{f} \times M_{2} \times w_{1}}{K_{f} \times 1000}$
$= \frac{1.5 \times 176 \times 75}{3.9 \times 1000}$

= 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.

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