At 100, ^oC the vapour pressure of a solution | vedclass

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(C) Given: $W_{B} = 6.5\, g$,$W_{A} = 100\, g$,$p_{s} = 732\, mm$,$K_{b} = 0.52$,$T_{b}^{o} = 100\, ^oC$,$p^{o} = 760\, mm$.Using Raoult's law for rel

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$101$

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(C) Given: $W_{B} = 6.5\, g$,$W_{A} = 100\, g$,$p_{s} = 732\, mm$,$K_{b} = 0.52$,$T_{b}^{o} = 100\, ^oC$,$p^{o} = 760\, mm$.Using Raoult's law for relative lowering of vapour pressure: $\frac{p^{o} - p_{s}}{p^{o}} = \frac{n_{2}}{n_{1}}$.$\frac{760 - 732}{760} = \frac{n_{2}}{100 / 18}$.$n_{2} = \frac{28 	imes 100}{760 	imes 18} \approx 0.2046\, mol$.Now,calculate elevation in boiling point: $\Delta T_{b} = K_{b} 	imes m = K_{b} 	imes \frac{n_{2} 	imes 1000}{W_{A} (g)}$.$\Delta T_{b} = \frac{0.52 	imes 0.2046 	imes 1000}{100} = 1.06\, ^oC$.Boiling point of solution $T_{b} = T_{b}^{o} + \Delta T_{b} = 100 + 1.06 = 101.06\, ^oC$.

At $100\, ^oC$ the vapour pressure of a solution of $6.5\, g$ of a solute in $100\, g$ water is $732\, mm.$ If $K_b = 0.52,$ the boiling point of this solution will be .........$^oC$

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