Ncert Exercises & Solutions

2.10 Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol⁻¹.

Here, elevation of boiling point ∆Tb = (100 + 273) − (99.63 + 273) = 0.37 K

Mass of water, w_l = 500 g

Molar mass of sucrose (C₁₂H₂₂O₁₁), M₂ = 11 × 12 + 22 × 1 + 11 × 16 = 342 g mol⁻¹

Molal elevation constant, K_b = 0.52 K kg mol⁻¹

We know that:

$∆ T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{l}}$
$\Rightarrow w_{2} = \frac{∆ T_{b} \times M_{2} \times w_{l}}{K_{b} \times 1000}$
$= \frac{0.37 \times 342 \times 500}{0.52 \times 1000}$

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

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