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Question

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


Solution

Given, $$\triangle T_b=100-99.63=0.37^o; K_b=0.52$$

Mass of water $$=500 \ g$$

Let $$x \ gm$$ of sucrose be added.

Molar mass of sucrose $$=342 \ g \ mol^{-1}$$

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

$$\triangle T_b = K_b \times m$$

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

$$x=121.67 \ gm$$

Therefore, $$121.67 \ gm$$ of sucrose must be added.

Chemistry

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