At higher altitude,the boiling point of water | vedclass

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(A) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^0 = 100\, ^oC - 95\, ^oC = 5\, K$. For $NaCl$,the van't Hoff factor $i = 1 + \a

Vedclass · Accepted Answer

$296.05$

Vedclass · Answer

(A) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^0 = 100\, ^oC - 95\, ^oC = 5\, K$. For $NaCl$,the van't Hoff factor $i = 1 + \alpha(n-1)$. Since $NaCl$ dissociates into $2$ ions $(n=2)$ and $\alpha = 0.9$,$i = 1 + 0.9(2-1) = 1.9$. The formula for elevation in boiling point is $\Delta T_b = i 	imes K_b 	imes m$,where $m$ is the molality. $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{w / 58.5}{1} = \frac{w}{58.5}$. Substituting the values: $5 = 1.9 	imes 0.52 	imes \frac{w}{58.5}$. $w = \frac{5 	imes 58.5}{1.9 	imes 0.52} = \frac{292.5}{0.988} \approx 296.05\, g$.

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

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