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JEE Advanced 2021 Paper 1

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The boiling point of water in a

0.1

molal silver nitrate solution (solution

A

) is

x^{\circ}\,\mathrm{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\mathrm{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\,\mathrm{K}\,\mathrm{kg}\,\mathrm{mol}^{-1}

; Boiling point of pure water as

100^{\circ}\,\mathrm{C}

Section: Chemistry

Question : 28 of 57

Marks: +1, -0

The value of

x

is___________

[JEE Adv 2021 P1]

Your Answer:

Validate

Solution:

In solution-A

\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{Ag}^+(\mathrm{aq}) + \mathrm{NO}_3^-(\mathrm{aq})

\because \Delta T_b = i k_b m

\Delta T_b^o = 2 \times 0.5 \times 0.1 \;\; [i=2]

(T - T^0) = 0.1

T_s^s = 100 + 0.1 = 100.1 = 0.1^{\circ} \mathrm{C} = 0.1 \mathrm{K}

In solution - B

\underset{0.05\,\text{m}}{\mathrm{AgNO}_3(\mathrm{aq})} \rightarrow \underset{0.05\,\text{m}}{\mathrm{Ag}^+(\mathrm{aq})} + \underset{0.05\,\text{m}}{\mathrm{NO}_3^-(\mathrm{aq})}

\underset{0.05\,\text{m}}{\mathrm{BaCl}_2(\mathrm{aq})} \rightarrow \underset{0.05\,\text{m}}{\mathrm{Ba}^{2+}(\mathrm{aq})} + 2\underset{0.01\,\text{m}}{\mathrm{Cl}^-(\mathrm{aq})}

\mathrm{Ag}^+

and

\mathrm{Cl}^-

combine to form

\mathrm{AgCl}

precipitate.

\mathrm{Ag}^+(\mathrm{aq}) + \mathrm{Cl}^-(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s})

t=0\;\;\;0.05\,\text{m}\;\;0.1\,\text{m}

t=\infty\;\;\;\;0\;\;\;\;\;\;0.05\,\text{m}

In final solution (B) total concentration of all ions is

[\mathrm{Cl}^-] + [\mathrm{NO}_3^-] + [\mathrm{Ba}^{2+}] = 0.05 + 0.05 + 0.05 = 0.15 \,\text{m}

\Delta T_b = 0.5 \times 0.15 = 0.075^{\circ} \mathrm{C}

B.P. of solution '

B

= 100.075^{\circ} \mathrm{C}

B.P. of solution 'A'

= 100.1^{\circ} \mathrm{C}

|y| = 100.1 - 100.075 = 0.025 = 2.5 \times 10^{-2}

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