Question

3.0 molal aqueous solution of an electrolyte $A_{2}B_{3}$ is 50 % ionised. The boiling point of the solution at $1$ $atm$ is: $\left[K_{b} \left(H_{2} O\right) = 0.52 \, K k g m o l^{- 1}\right]$

• A. 274.76 K
• B. 377.68 K
• C. 374.68 K
• D. 104.68 K

Answer 1

Answer: (B)

$A_{2}B_{3}\left(\right.aq\left.\right) \rightarrow 2\left(A^{3}\right)^{+}\left(\right.aq\left.\right)+3B^{2 -}\left(\right.aq\left.\right)$

$n=5$

$\Delta T_{b}=i.K_{b}.m$

$=\left[\right.1+\left(\right.n-1\left.\right)\alpha \left]\right.K_{b}.m.$

$=\left[\right.1+4\alpha \left]\right.K_{b}.m.$

$=\left[\right.1+4\left(\right.0.5\left.\right)\left]\right.0.52\times 3$

$=\left(\right.1+2\left.\right)\left(\right.0.52\left.\right)\times 3$

$=4.68$

$A_{b}=373+4.68=377.68K$