AB2 is 10% dissociated in water
Question:

$\mathrm{AB}_{2}$ is $10 \%$ dissociated in water to $\mathrm{A}^{2+}$ and $\mathrm{B}^{-}$. The boiling point of a $10.0$ molal aqueous

solution of $\mathrm{AB}_{2}$ is ${ }^{\circ} \mathrm{C}$. (Round off to the Nearest Integer).

[Given : Molal elevation constant of water $\mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ boiling point of pure water $\left.=100^{\circ} \mathrm{C}\right]$

Solution:

$\mathrm{AB}_{2} \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}$

$\begin{array}{llll}\mathrm{t}=0 & \mathrm{a} & 0 & 0 \\ \mathrm{t}=\mathrm{t} & \mathrm{a}-\mathrm{a} \alpha & \mathrm{a} \alpha & 2 \mathrm{a} \alpha\end{array}$

$\mathrm{n}_{\mathrm{T}}=\mathrm{a}-\mathrm{a} \alpha+\mathrm{a} \alpha+2 \mathrm{a} \alpha$

$=a(1+2 \alpha)$

so $i=1+2 \alpha$

Now $\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{m} \times \mathrm{K}_{\mathrm{b}}$

$\Delta \mathrm{T}_{\mathrm{b}}=(1+2 \alpha) \times \mathrm{m} \times \mathrm{K}_{\mathrm{b}}$

$\alpha=0.1 \quad \mathrm{~m}=10 \quad \mathrm{~K}_{\mathrm{b}}=0.5$

$\Delta \mathrm{T}_{\mathrm{h}}=1.2 \times 10 \times 0.5$

= 6

So boiling point = 106

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