JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Question Stem
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 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.
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
