Boiling point of water at 750 mm Hg 99.63⁰C. How much sucrose is to be added to 500g of water such that it boils at 100⁰C.

Solution

Solution:
here, elevation of boiling point

∆ T_{b}

= (100+273)-(99.63+273) = 0.37K
mass of water, w₁ = 500g
Boiling point of water at 750 mm Hg = 99.63⁰C

M_{B} = \frac{k_{b} \times 1000 \times ω_{B}}{ω_{A} \times ∆ T_{b}}

k_{b} = 0.52 k kg {mol}^{- 1}, ω_{B} = 500 g ∆ T_{b} = 100 - 99.63 = 0.37 ° C

Molar mass of sucrose,

C_{12} H_{22} O_{11} (M_{g})

= 12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342 g {mol}^{- 1}

342 = \frac{0.52 \times 1000 \times W_{B}}{500 \times 0.37}

W_{B} = \frac{342 \times 500 \times 0.37}{0.52 \times 1000} = 121.67 g

hence , 121.67 g of sucrose is to obtained.

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