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Q. If 0.5 g of a solute (molar mass 100 g $ mo{{l}^{-1}} $ ) in of solvent elevates the boiling point by 1K, the molar boiling point constant of the solvent is

CMC MedicalCMC Medical 2008

Solution:

By using the relation $ \Delta {{T}_{b}}=\frac{{{k}_{b}}\times {{w}_{B}}\times 1000}{{{M}_{B}}\times {{w}_{A}}} $ Here, $ \Delta {{T}_{b}}=1\,K, $ $ {{w}_{B}}=0.5g, $ $ {{w}_{A}}=25g $ $ {{M}_{B}}=100\,g\,\text{mo}{{\text{l}}^{-1}} $ $ \therefore $ $ 1=\frac{{{k}_{b}}\times 0.5\times 1000}{100\times 25} $ $ {{k}_{b}}=5 $