JEE Advanced · Chemistry · 16. Solutions

The boiling point of water in a $0.1$ molal silver nitrate solution (solution $\mathbf{A}$ ) is $\mathbf{x}^{\circ} \mathrm{C}$. To this solution $\mathbf{A}$, 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 $\mathbf{A}$ and $\mathbf{B}$ is $\mathbf{y} \times 10^{-2}{ }^{\circ} \mathrm{C}$.
(Assume: Densities of the solutions $\mathbf{A}$ and $\mathbf{B}$ are the same as that of water and the soluble salts dissociate completely.
Use: Molal elevation constant (Ebullioscopic Constant), $K_{b}=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$; Boiling point of pure water as $100^{\circ} \mathrm{C}$.)
The value of $|y|$ is ___.

A 2.5
B 3.5
C 4.5
D 5.5

Verified Solution

Answer & Solution

Correct Answer

(A) 2.5

Step-by-step Solution

Detailed explanation

After adding

{BaCl}_{2}

solution in

{AgNO}_{3}

solution

AgCl

will be precipitate. Since volume of both solutions are equal let's take

V = 1 L

of each.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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From JEE Advanced

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

If a tangent to a suitable conic (Column 1) is found to be

y = x + 8

and its point of contact is

(8, 16)

, then which of the following options is the only Correct combination?

JEE Advanced 2017 Hard

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