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Solutions question

2021 · Shift 1 · Q9
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Solutions question

2021 · Shift 1 · Q9

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 x is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 100.1

  1. Use elevation in boiling point formula

    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​=0.5 K kg mol−1
    • m=0.1m = 0.1m=0.1 molal

  2. For solution A: 0.10.10.1 molal AgNO3\text{AgNO}_3AgNO3​

    Silver nitrate dissociates completely as: AgNO3→Ag++NO3−\text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^-AgNO3​→Ag++NO3−​

    So, i=2i = 2i=2

  3. Calculate boiling point elevation

    ΔTb=iKbm=2×0.5×0.1=0.1∘C\Delta T_b = iK_bm = 2 \times 0.5 \times 0.1 = 0.1^\circ \text{C}ΔTb​=iKb​m=2×0.5×0.1=0.1∘C

  4. Boiling point of solution A

    Since pure water boils at 100∘C100^\circ \text{C}100∘C,

    x=100+0.1=100.1∘Cx = 100 + 0.1 = 100.1^\circ \text{C}x=100+0.1=100.1∘C

  5. Compare with stored answer

    Derived value: x=100.1x = 100.1x=100.1

    Stored correct answer: 100.1100.1100.1

    They agree.

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