Boiling point of water at 750 mm Hg is 99.63°C.
Question:

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.

Solution:

Here, elevation of boiling point ΔTb = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water, wl = 500 g

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

= 342 g mol−1

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

We know that:

$\Delta T_{b}=\frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$

$\Rightarrow w_{2}=\frac{\Delta T_{b} \times M_{2} \times w_{1}}{K_{b} \times 1000}$

$=\frac{0.37 \times 342 \times 500}{0.52 \times 1000}$

= 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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