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Solutions question

2021 · Shift 1 · Q10
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Solutions question

2021 · Shift 1 · Q10

JEE AdvancedChemistrySolutionsNumerical+2 / −1
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 is y ×\times× 10 −-− 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), Kb = 0.5 K kg mol −-− 1; Boiling point of pure water as 100 ∘{}^\circ∘ C.)The value of | y | is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2.5

  1. Boiling point elevation formula

For a solution,

ΔTb=iKbm\Delta T_b = i K_b mΔTb​=iKb​m

where:

  • iii = van't Hoff factor
  • Kb=0.5 K kg mol−1K_b = 0.5\,\text{K kg mol}^{-1}Kb​=0.5K kg mol−1
  • mmm = molality
  1. Solution A: 0.1 molal AgNO3\text{AgNO}_3AgNO3​

Since AgNO3\text{AgNO}_3AgNO3​ dissociates completely:

AgNO3→Ag++NO3−\text{AgNO}_3 \to \text{Ag}^+ + \text{NO}_3^-AgNO3​→Ag++NO3−​

So,

i=2i = 2i=2

Hence,

ΔTb(A)=2×0.5×0.1=0.1∘C\Delta T_b(A) = 2 \times 0.5 \times 0.1 = 0.1^\circ \text{C}ΔTb​(A)=2×0.5×0.1=0.1∘C

Therefore boiling point of solution A is

x=100+0.1=100.1∘Cx = 100 + 0.1 = 100.1^\circ \text{C}x=100+0.1=100.1∘C
  1. Take convenient volumes using density = density of water

Let us take 1 L of solution A.

Since density is same as water, mass of 1 L solution ≈1 kg\approx 1\,\text{kg}≈1kg. For a dilute solution, 0.1 molal means approximately 0.1 mol solute in 1 kg solvent. Using the given approximation, in 1 L solution we take:

  • moles of AgNO3=0.1\text{AgNO}_3 = 0.1AgNO3​=0.1

Similarly, take 1 L of 0.1 molal BaCl2\text{BaCl}_2BaCl2​ solution. Then:

  • moles of BaCl2=0.1\text{BaCl}_2 = 0.1BaCl2​=0.1

After mixing equal volumes, total volume =2= 2=2 L, hence mass of solution ≈2\approx 2≈2 kg. So effective molality of each solute in mixture B is approximately

0.12=0.05\frac{0.1}{2} = 0.0520.1​=0.05
  1. Boiling point elevation in solution B

For AgNO3\text{AgNO}_3AgNO3​, i=2i=2i=2:

ΔTb(from AgNO3)=2×0.5×0.05=0.05∘C\Delta T_b(\text{from AgNO}_3) = 2 \times 0.5 \times 0.05 = 0.05^\circ \text{C}ΔTb​(from AgNO3​)=2×0.5×0.05=0.05∘C

For BaCl2\text{BaCl}_2BaCl2​, complete dissociation:

BaCl2→Ba2++2Cl−\text{BaCl}_2 \to \text{Ba}^{2+} + 2\text{Cl}^-BaCl2​→Ba2++2Cl−

So,

i=3i = 3i=3

Hence,

ΔTb(from BaCl2)=3×0.5×0.05=0.075∘C\Delta T_b(\text{from BaCl}_2) = 3 \times 0.5 \times 0.05 = 0.075^\circ \text{C}ΔTb​(from BaCl2​)=3×0.5×0.05=0.075∘C

Total elevation in B:

ΔTb(B)=0.05+0.075=0.125∘C\Delta T_b(B) = 0.05 + 0.075 = 0.125^\circ \text{C}ΔTb​(B)=0.05+0.075=0.125∘C

Thus boiling point of B is

100+0.125=100.125∘C100 + 0.125 = 100.125^\circ \text{C}100+0.125=100.125∘C
  1. Difference in boiling points
∣Tb(B)−Tb(A)∣=∣100.125−100.1∣=0.025∘C|T_b(B)-T_b(A)| = |100.125 - 100.1| = 0.025^\circ \text{C}∣Tb​(B)−Tb​(A)∣=∣100.125−100.1∣=0.025∘C

Given this is written as

y×10−2 ∘Cy \times 10^{-2}\,{}^\circ\text{C}y×10−2∘C

so

0.025=y×10−20.025 = y \times 10^{-2}0.025=y×10−2 y=2.5y = 2.5y=2.5

Therefore,

∣y∣=2.5|y| = 2.5∣y∣=2.5
  1. Comparison with stored answer

Derived answer: 2.52.52.5

Stored correct answer: 2.52.52.5

They match.

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