#### 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 1

Here, elevation of boiling point Δ*T*_{b} = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water, *w*_{l} = 500 g

Molar mass of sucrose (C_{12}H_{22}O_{11}), *M*_{2} = 11 × 12 + 22 × 1 + 11 × 16

= 342 g mol^{−1}

Molal elevation constant, *K*_{b} = 0.52 K kg mol^{−1}

We know that:

`triangleT_b = (K_bxx1000xxw_2)/(M_2xxw_1)`

`=>w_2 = (triangleT_bxxM_2xxw_1)/(K_bxx1000)`

`= (0.37xx342xx500)/(0.52xx1000)`

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.

#### Solution 2

Given ΔT_{b} = 100 - 96.63 = 3.37º

Mass of water, w_{1} = 500g

Molar mass of water, M_{1} = 18g mol^{-1}

Molar mass of sucrose, M_{2} = 342g mol^{-1}

To find :Mass of sucrose, w_{2} = ?

Solution: we know, ΔT_{b }= K_{b} x m

= `K_bxxw_2/M_2xx1000/w_1`

`=>w_2 = (M_2xxw_1xxtriangleT_B)/(1000xxK_b)` = `(342xx500xx3.37)/(1000xx0.52)`

w_{2} = 1108.2g

Mass of solute w_{2} = 1.11 kg

#### Solution 3

Here, elevation of boiling point Δ*T*_{b} = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water, *w*_{l} = 500 g

Molar mass of sucrose (C_{12}H_{22}O_{11}), *M*_{2} = 11 × 12 + 22 × 1 + 11 × 16

= 342 g mol^{−1}

Molal elevation constant, *K*_{b} = 0.52 K kg mol^{−1}

We know that:

`triangleT_b = (K_bxx1000xxw_2)/(M_2xxw_1)`

`=>w_2 = (triangleT_bxxM_2xxw_1)/(K_bxx1000)`

`= (0.37xx342xx500)/(0.52xx1000)`

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.