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Q. Addition of 1.0 g of a compound to 10 g of water .increases the boiling point by 0.3?C. The amount of compound needed to prepare a 500 mL of 0.1 solution is [Given: assume negligible dissociation or association of the compound, boiling point constant $ {{k}_{b}} $ of water $ =0.513kg\,mo{{l}^{-1}} $ ]

VMMC MedicalVMMC Medical 2015

Solution:

$ \Delta {{T}_{b}}={{m}_{solute}}{{K}_{b}} $ $ =0.3=\frac{\left( \frac{1.0}{M} \right)\times 1000}{10}\times 0.513 $ = 171 [M = molecular weight of solute] We have, Molarity $ \text{=}\frac{\frac{\text{Weight}}{\text{M}}\text{ }\!\!\times\!\!\text{ 1000}}{\text{Volume}\,\text{of}\,\text{solution(in}\,\text{mL)}} $ $ \text{0}\text{.1=}\frac{\frac{\text{Weight}}{171}\text{ }\!\!\times\!\!\text{ 1000}}{500} $ Weight = 5.55 g