80 mole percent of MgCl_2 is dissociated in a | vedclass

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(B) $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$Initial moles: $1$,$0$,$0$At equilibrium: $1-\alpha$,$\alpha$,$2\alpha$Total particles $i = 1 + 2\alpha$. Give

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$48$

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(B) $MgCl_2 ightarrow Mg^{2+} + 2Cl^-$Initial moles: $1$,$0$,$0$At equilibrium: $1-\alpha$,$\alpha$,$2\alpha$Total particles $i = 1 + 2\alpha$. Given $\alpha = 0.8$,so $i = 1 + 2(0.8) = 2.6$.For a $1.0$ molal solution,$n_2 = 1$ mole of solute in $1000 \ g$ of water ($n_1 = 1000/18 = 55.55$ moles).Using Raoult's Law: $\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{i 	imes n_2}{n_1 + i 	imes n_2} \approx \frac{i 	imes n_2}{n_1}$.$\frac{50 - p_s}{50} = \frac{2.6 	imes 1}{55.55} = 0.0468$.$50 - p_s = 2.34 \implies p_s = 47.66 \ mm \ Hg$.Rounding to the nearest integer,$p_s \approx 48 \ mm \ Hg$.

$80$ mole percent of $MgCl_2$ is dissociated in aqueous solution. The vapour pressure of $1.0$ molal aqueous solution of $MgCl_2$ at $38^{\circ} C$ is $.........$ $mm$ $Hg$. (Nearest integer)
Given : Vapour pressure of water at $38^{\circ} C$ is $50$ $mm$ $Hg$.

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$80$ mole percent of $MgCl_2$ is dissociated in aqueous solution. The vapour pressure of $1.0$ molal aqueous solution of $MgCl_2$ at $38^{\circ} C$ is $.........$ $mm$ $Hg$. (Nearest integer)Given : Vapour pressure of water at $38^{\circ} C$ is $50$ $mm$ $Hg$.