Question

$Mg(OH{)}_2$ is partially soluble in water and maximum $pH$ of aqueous solution of $Mg(OH{)}_2$ at ${100}^{o}C$ is 11. The boiling point of aqueous $Mg(OH{)}_2$ solution is [given $K_w$ of water at ${100}^{o}C$ is ${10}^{-13}$ and $(K_b{)}_{H_{2}O}$ is $0.52K{m}^{-1}$ assume $m=M$, boiling point of water is ${100}^{o}C$]

• A. ${100.0078}^{o}C$
• B. ${101.0026}^{o}C$
• C. ${101.26}^{o}C$
• D. ${100.26}^{o}C$

Answer 1

The correct option is A ${100.0078}^{o}C$

$pH=11$
$\therefore$ $poH=2$ [∵ $k_w={10}^{-13}$]
$\therefore$ $O{H}^{-}={10}^{-2}$
$\therefore$ $M{g}^{2+}=\frac{{10}^2}{2}=5\times {10}^{-3}$

and $i=3$

$\therefore$ $\Delta T_b=i{K}_{b}M=3\left(0.52\right)(5\times {10}^{-3})$
$=0.0078$

$\therefore$ $T_b=100+0.0078={100.0078}^{o}C$