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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.)

Question 47

The value of x is ___.


Correct Answer: 100.10 - 100.10

The boiling‐point elevation formula is
$$\Delta T_b = i\,K_b\,m$$
where $$i$$ is the van’t Hoff factor, $$K_b$$ is the ebullioscopic constant and $$m$$ is the molality.

For solution A:
• Solute = $$AgNO_3$$ which dissociates completely as $$AgNO_3 \rightarrow Ag^{+} + NO_3^{-}$$.
• Number of ions produced = 2, so $$i = 2$$.
• Given molality $$m = 0.1\;{\rm mol\,kg^{-1}}$$.
• Given $$K_b = 0.5\;{\rm K\,kg\,mol^{-1}}$$.

Substituting in the formula:
$$\Delta T_b = 2 \times 0.5 \times 0.1 = 0.1\;{\rm K}$$.

The boiling point of pure water is $$100.00^\circ{\rm C}$$, therefore the boiling point of solution A is
$$x = 100.00^\circ{\rm C} + 0.10^\circ{\rm C} = 100.10^\circ{\rm C}$$.

Hence, the value of $$x$$ is 100.10 °C.

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