Boiling point of water at $750 m m H g$ is ${99.63}^{\circ} C$ . How much sucrose is to be added to $500 g$ of water such that it boils at $100^{\circ} C$ .

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Solution

Given, $△ T_{b} = 100 - 99.63 = {0.37}^{o}; K_{b} = 0.52$

Mass of water $= 500 g$

Let $x g m$ of sucrose be added.

Molar mass of sucrose $= 342 g m o l^{- 1}$

Molality $(M) = \frac{x \times 1000}{342 \times 500} = \frac{x}{171}$

$△ T_{b} = K_{b} \times m$

$0.37 = 0.52 \times \frac{x}{171}$

$x = 121.67 g m$

Therefore, $121.67 g m$ of sucrose must be added.

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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.

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