At higher altitude,the boiling point of water | vedclass

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Question

(A) The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$,where $m$ is the molality.Given: $\Delta T_b = 100\,^oC - 95\,^oC

Vedclass · Accepted Answer

$296$

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(A) The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$,where $m$ is the molality.Given: $\Delta T_b = 100\,^oC - 95\,^oC = 5\,K$,$K_b = 0.52\,K\,kg\,mol^{-1}$,$W_{solvent} = 1\,kg$,$M_{NaCl} = 58.5\,g\,mol^{-1}$.For $NaCl$,$n = 2$. With $\alpha = 0.9$,the van't Hoff factor $i = 1 + \alpha(n-1) = 1 + 0.9(2-1) = 1.9$.Using the formula $\Delta T_b = i 	imes K_b 	imes \frac{w}{M 	imes W_{solvent}}$:$5 = 1.9 	imes 0.52 	imes \frac{w}{58.5 	imes 1}$.$w = \frac{5 	imes 58.5}{1.9 	imes 0.52} = \frac{292.5}{0.988} \approx 296.05\,g$.Thus,the amount of $NaCl$ required is approximately $296\,g$.

At higher altitude,the boiling point of water is $95\,^oC$. The amount of $NaCl$ added to $1\,kg$ of water $(K_b = 0.52\,K\,kg\,mol^{-1})$ in order to raise the boiling point of the solution to $100\,^oC$ (assume $90\%$ ionisation of $NaCl$) is .......... $g$.

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