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), $K _{ b }=0.5 \,K \,kg\, mol ^{-1}$;
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 $BaCl _{2}$.

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 ' $B$ '
$=100.10-100.075=0.025= y \times 10^{-2}$ (given)
So, $y =2.5$