The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is . 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 and is .
(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), ; Boiling point of pure water as 100°C.)
The value of is ____.
(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), ; Boiling point of pure water as 100°C.)
The value of is ____.
Answer: 2.5
Solution
When equal volumes are mixed, molalities are halved:
From AgNO₃: [Ag⁺] = 0.05 m, [NO₃⁻] = 0.05 m
From BaCl₂: [Ba²⁺] = 0.05 m, [Cl⁻] = 0.1 m
Ag⁺ + Cl⁻ → AgCl(s) precipitates
0.05 m Ag⁺ reacts with 0.05 m Cl⁻, leaving 0.05 m Cl⁻ in solution
Total ions in solution B: [Cl⁻] + [NO₃⁻] + [Ba²⁺] = 0.05 + 0.05 + 0.05 = 0.15 m
for B = 0.5 × 0.15 = 0.075°C
Boiling point of B = 100.075°C
Difference = 100.10 - 100.075 = 0.025°C = 2.5 × 10⁻²°C
Therefore, |y| = 2.50
From AgNO₃: [Ag⁺] = 0.05 m, [NO₃⁻] = 0.05 m
From BaCl₂: [Ba²⁺] = 0.05 m, [Cl⁻] = 0.1 m
Ag⁺ + Cl⁻ → AgCl(s) precipitates
0.05 m Ag⁺ reacts with 0.05 m Cl⁻, leaving 0.05 m Cl⁻ in solution
Total ions in solution B: [Cl⁻] + [NO₃⁻] + [Ba²⁺] = 0.05 + 0.05 + 0.05 = 0.15 m
for B = 0.5 × 0.15 = 0.075°C
Boiling point of B = 100.075°C
Difference = 100.10 - 100.075 = 0.025°C = 2.5 × 10⁻²°C
Therefore, |y| = 2.50
JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Numerical answer