At a certain temperature,the vapour pressure | vedclass

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(B) Given: $P^0 = 76 \ mm \ Hg$,$P_s = 38 \ mm \ Hg$. Moles of water $(n_A)$ = $\frac{36 \ g}{18 \ g/mol} = 2 \ mol$. Moles of solute $(n_B)$ = $1 \ m

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$50$

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(B) Given: $P^0 = 76 \ mm \ Hg$,$P_s = 38 \ mm \ Hg$. Moles of water $(n_A)$ = $\frac{36 \ g}{18 \ g/mol} = 2 \ mol$. Moles of solute $(n_B)$ = $1 \ mol$. Using Raoult's law for non-volatile solute: $\frac{P^0 - P_s}{P_s} = i 	imes \frac{n_B}{n_A}$. $\frac{76 - 38}{38} = i 	imes \frac{1}{2} \implies 1 = i 	imes \frac{1}{2} \implies i = 2$. For $Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-}$,the number of ions produced per formula unit is $n = 3$. $i = 1 + (n - 1)\alpha \implies 2 = 1 + (3 - 1)\alpha$. $2 = 1 + 2\alpha \implies 2\alpha = 1 \implies \alpha = 0.5$. Percentage dissociation = $0.5 	imes 100 = 50\%$.

At a certain temperature,the vapour pressure of pure water is $76 \ mm$ of $Hg$. When $1 \ mol$ of $Na_2SO_4$ is dissolved in $36 \ mL$ of water,the vapour pressure of the solution becomes $38 \ mm$ of $Hg$. What is the percentage dissociation of $Na_2SO_4$ in the solution?

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