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The rise in boiling point of a solution containing \(1.8 \mathrm{~g}\) of glucose in \(100 \mathrm{~g}\) of solvent is \(0.1^{\circ} \mathrm{C}\). The molal elevation constant of the liquid is

A \(1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
B \(2 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
C \(10 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
D \(0.1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)

Step-by-step Solution

Detailed explanation

Given, amount of glucose \((w)=1.8 \mathrm{~g}\)
Amount of solvent \((W)=100 \mathrm{~g}\)
\(\Delta T_b=0.1^{\circ} \mathrm{C}\)
Molecular mass of glucose \(=180\)
Molal elevation constant of the liquid, \(\Delta T_b=0.1^{\circ} \mathrm{C}\)
We know that,
\[
\begin{aligned}
K_b & =\frac{\Delta T_b \times m \times W}{1000 \times w} \\
& =\frac{0.1 \times 180 \times 100}{1000 \times 1.8}=1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}
\end{aligned}
\]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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