# Colligative properties, Elevation of boling point (EBP)

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# At  ${100}^{0}\mathrm{C}$   the vapour pressure of a solution of  $6.5\mathrm{g}$ of a solute in $100\mathrm{g}$ water is $732\mathrm{mm}$. If ${\mathrm{k}}_{\mathrm{b}}=0.52$ the boiling point of this solution will be

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Solution

## $\frac{\mathrm{p}°-{\mathrm{p}}_{\mathrm{s}}}{\mathrm{p}°}=\frac{{\mathrm{n}}_{2}}{{\mathrm{n}}_{1}}⇒\frac{760-732}{760}=\frac{{\mathrm{n}}_{2}}{100/18}$${\mathrm{\Delta T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}×\mathrm{m}$${\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}°={\mathrm{K}}_{\mathrm{b}}×\frac{{\mathrm{n}}_{2}×1000}{{\mathrm{W}}_{\mathrm{A}}\left(\mathrm{g}\right)}$$\begin{array}{l}{\mathrm{T}}_{\mathrm{b}}-100°\mathrm{C}=\frac{0.52×0.2046×1000}{100}=1.06\\ {\mathrm{T}}_{\mathrm{b}}=100+1.06=101.06°\mathrm{C}\end{array}$

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