At higher altitude,the boiling point of water | vedclass

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(A) The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$. Here,$\Delta T_b = 100^{\circ}C - 95^{\circ}C = 5\,K$. For $NaCl$

Vedclass · Accepted Answer

$296.05$

Vedclass · Answer

(A) The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$. Here,$\Delta T_b = 100^{\circ}C - 95^{\circ}C = 5\,K$. For $NaCl$,the van't Hoff factor $i = 1 + \alpha(n-1)$. With $\alpha = 0.9$ and $n = 2$,$i = 1 + 0.9(2-1) = 1.9$. Molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$,where $w_2$ is the mass of $NaCl$ and $M_2 = 58.5\,g/mol$. Substituting the values: $5 = 1.9 	imes 0.52 	imes \frac{w_2}{58.5 	imes 1}$. $w_2 = \frac{5 	imes 58.5}{1.9 	imes 0.52} = \frac{292.5}{0.988} \approx 296.05\,g$.

At higher altitude,the boiling point of water is $95\,^{\circ}C$. 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 $100\,^{\circ}C$ (assuming $90\%$ ionization of $NaCl$) is ........... $g$.

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