Aqueous solution of glucose is $35 % w / w$ . Then calculate the boiling point of the solution.
Given: Aqueous solution of 1 molal urea solution boils at ${100.60}^{\circ} C$ and water boils at 1 atm pressure.

{100.01}^{\circ} C

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{103.41}^{\circ} C

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{97.83}^{\circ} C

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{101.79}^{\circ} C

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Solution

The correct option is D ${101.79}^{\circ} C$
Water boils at 1 atmospheric pressure.

Thus, boiling point of $H_{2} O = 100^{\circ} C$

1 molal urea solution boils at ${100.6}^{\circ} C$

Elevation in boling point of urea solution:

$∴ Δ T_{b} = (100.6 - 100)^{\circ} C = {0.60}^{\circ} C$

$Δ T_{b} = K_{b} \times m o l a l i t y$

$0.60 = K_{b} \times 1$

$∴ K_{b} (H_{2} O) = 0.60 K m o l^{- 1} k g$

Glucose solution is 35% by weight of solution.

Thus,
$w_{1} (g l u c o s e) = 35 g$

$w_{2} (s o l v e n t H_{2} O) = 65 g$

$Molar mass of glucose (m_{1}) = 180 g$

Elevation in boiling point for glucose solution:

$Δ T_{b} = \frac{1000 \times K_{b} \times w_{1}}{m_{1} \times w_{2}}$

$Δ T_{b} = \frac{1000 \times 0.60 \times 35}{180 \times 65} = {1.79}^{\circ} C$

$∴ Boiling point of glucose solution = 100^{\circ} C + {1.79}^{\circ} C = {101.79}^{\circ} C$

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Aqueous solution of glucose is 35% w/w. Then calculate the boiling point of the solution. Given: Aqueous solution of 1 molal urea solution boils at 100.60∘C and water boils at 1 atm pressure.

Aqueous solution of glucose is $35 % w / w$ . Then calculate the boiling point of the solution.
Given: Aqueous solution of 1 molal urea solution boils at ${100.60}^{\circ} C$ and water boils at 1 atm pressure.