Calculate the boiling point of an aqueous solution containing 18 g glucose in 100 g water if the molal elevation constant of water is $0.5 \text{K kg mol}^{-1}$.
[Molar mass of glucose $= 180 \text{g mol}^{-1}$ and boiling point of water $= 100 ^{\circ}\text{C}$]

Question

What is the depression in the freezing point?

Answer 1

Step 1: Molality:
$m = \dfrac{18/180}{100/1000} = \dfrac{0.1}{0.1} = 1$ mol/kg.

Step 2: Elevation:
$\Delta T_b = 0.5\times1 = 0.5$ K.

Step 3: Final boiling point:
$T_b = 100 + 0.5 = 100.5\ ^{\circ}$C, option (B). Glucose does not ionise, so no van't Hoff factor is needed.

Final Answer:
The boiling point is 100.5 C. \[ \boxed{\text{(B) }100.5\ ^{\circ}\text{C}} \]

Calculate the boiling point of an aqueous solution containing 18 g glucose in 100 g water if the molal elevation constant of water is \(0.5 \text{K kg mol}^{-1}\).
[Molar mass of glucose \(= 180 \text{g mol}^{-1}\) and boiling point of water \(= 100 ^{\circ}\text{C}\)]

Show Hint

The Correct Option is B

Solution and Explanation

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Calculate the boiling point of an aqueous solution containing 18 g glucose in 100 g water if the molal elevation constant of water is \(0.5 \text{K kg mol}^{-1}\).[Molar mass of glucose \(= 180 \text{g mol}^{-1}\) and boiling point of water \(= 100 ^{\circ}\text{C}\)]