The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is $\mathbf{x}^{\circ}\mathrm{C}$. 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 $\mathbf{A}$ and $\mathbf{B}$ is $\mathbf{y} \times 10^{-2}{}^{\circ}\mathrm{C}$. (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), $\mathrm{K}_{\mathrm{b}}=0.5~\mathrm{K~kg~mol}^{-1}$; Boiling point of pure water as 100°C.) The value of $\mathbf{x}$ is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Solutions

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

y \times 1 0^{- 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°C.)

The value of

x

is ____.

Answer: 100.1

Solution

AgNO₃ dissociates completely: AgNO₃ → Ag⁺ + NO₃⁻

For 0.1 molal AgNO₃, total molality of particles = 0.1 + 0.1 = 0.2 m

Boiling point elevation:

Δ T_{b} = K_{b} \times m = 0.5 \times 0.2 = 0.1

K = 0.1°C

Boiling point of solution A:

x = 100 + 0.1 = 100.10

°C

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Numerical answer

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Solutions

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