Question

The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is $x^{\circ }C.$ To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is $y\times {10}^{-2}$ ${}^{\circ }C$
(Assume : Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely.
Use : Molal elevation constant (Ebullioscopic Constant), $K_b=0.5Kkgmo{l}^{-1};$
Boiling point of pure water as ${100}^{\circ }C.$)

The value of |$y$| is ___.

Answer 1

Let solution ‘B’ is prepared by mixing 1 L (=1000 g) of solution ‘A’ with 1 L (= 1000 g) of solution of $BaC{l}_2.$
$\begin{array}{ccccc}& BaC{l}_2+& 2AgN{O}_3\to & 2AgCl\left(s\right)+& Ba(N{O}_3{)}_2\\ \text{initial mole}& 0.1& 0.1& -& 0\\ \text{Final moles}& 0.05& 0& -& 0.05\end{array}$
The $i$ value of both $BaC{l}_2$ and $Ba(N{O}_3{)}_2$ is $3$
So, molality of new solution $\left(\frac{i_1\times m_1+i_2\times m_2}{2}\right)$
$\left(\frac{3\times 0.05+3\times 0.05}{2}\right)$
$=0.15$
Now, Elevation of boiling point of solution 'B' be $\left(\Delta T{{}_b}^1\right)$
$\Delta T_b^1=0.15\times k_b$
$=0.15\times \frac{1}{2}$
$=0.075$
Now, $T{{}_b}^1={100.075}^{\circ }C$
So, difference of boiling point of ‘A’ and ‘B’ $=100.10–100.075=0.025=2.5\times {10}^{-2}$
So, $y=2.5$