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Question

Question: The rise in the boiling point of a solution containing 1.8 gram of glucose in \(100g\) of a solvent ...

The rise in the boiling point of a solution containing 1.8 gram of glucose in 100g100g of a solvent in 0.1oC0.1^{o}C. The molal elevation constant of the liquid is.

A

0.01 K/mK/m

B

0.1K/m0.1K/m

C

1K/m1K/m

D

10K/m10K/m

Answer

1K/m1K/m

Explanation

Solution

Kb=ΔTbm=0.1×1001.8180×1000=1K/mK_{b} = \frac{\Delta T_{b}}{m} = \frac{0.1 \times 100}{\frac{1.8}{180} \times 1000} = 1K/m.