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Q. If the amount of a non-electrolyte dissolved is doubled but that of solvent is quadrupled, the elevation in boiling point of the solution will be

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Solution:

$\Delta T_{b}=\frac{1000 K_{b} w_{2}}{w_{1} \times M_{2}}$

New $\Delta T_{b}'=\frac{1000 K_{b}\left(2 w_{2}\right)}{\left(4 w_{1}\right) \times M_{2}}=\frac{1}{2}\left(\frac{1000 K_{b} w_{2}}{w_{1} M_{2}}\right) $

$\therefore \Delta T_{b}'=\frac{\Delta T_{b}}{2}$