The boiling point of water in a 0.1 molal silver nitrate solution (solution A ) is x ° 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 × 10 -20 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 ° C .) The value of y is \!\! \!\! .

Chemistry · Solutions and Colligative Properties

JEE Advanced 2021 — Paper 1 — Question 43

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 $\mathbf{B}$ .

The difference in the boiling points of water in the two solutions $\mathbf{A}$ and $\mathbf{B}$ is $\mathbf{y} \times 10^{-20} \mathrm{C}$ .

(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), $\mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ;

Boiling point of pure water as $100^{\circ} \mathrm{C}$ .)

The value of $\left| \mathbf{y} \right|$ is $\text{ }\!\!~\!\!\text{ }$ .

Answer: 2.5

Numerical answer — enter this value.

Step-by-step solution

Let solution ' $B$ ' is prepared by mixing $1\text{ }\!\!~\!\!\text{ L}\left( =1000\text{ }\!\!~\!\!\text{ g} \right)$ of solution 'A' with $1\text{ }\!\!~\!\!\text{ L}\left( =1000\text{ }\!\!~\!\!\text{ g} \right)$ of solution of $\text{BaC}{{\text{l}}_{2}}$ .

Initial moles

$\text{BaC}{{\text{l}}_{2}}+2\text{AgN}{{\text{O}}_{3}}2\text{AgCl}\left( \text{ }\!\!~\!\!\text{ s} \right)+\text{Ba}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}$

Moles after reaction 0.1-0.05

$\begin{array}{*{35}{l}}=0.05\text{ }\!\!~\!\!\text{ moles }\!\!~\!\!\text{ } & 0 & - & 0.05\text{ }\!\!~\!\!\text{ moles }\!\!~\!\!\text{ } \\\end{array}$

So, molality of new solution $=\left( \frac{{{i}_{1}}\times {{m}_{1}}+{{i}_{2}}\times {{m}_{2}}}{2} \right)$

$\begin{array}{*{35}{r}}{} & ~=\left( \frac{3\times 0.05+3\times 0.05}{2} \right) \\{} & ~=0.15 \\\end{array}$

Now, Elevation of boiling point of solution 'B' be $\left( \text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}{{~}^{1}} \right)$ $\text{ }\!\!\Delta\!\!\text{ T}_{\text{b}}^{1}=0.15\times {{\text{k}}_{\text{b}}}$ $=0.15\times \frac{1}{2}$ $=0.075$ Now, ${{\text{T}}_{\text{b}}}{{~}^{1}}=100.075{{~}^{\circ }}\text{C}$ So, difference of boiling point of ' A ' and ' B ' $=100.10-100.075=0.025=\text{y}\times {{10}^{-2}}$ (given) So, $y=2.5$

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Exam: JEE Advanced 2021
Paper: Paper 1
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Abnormal Colligative Properties - van't Hoff Factor

The value of x is .

Answer: 2.5

Step-by-step solution

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