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ChemMedAdv 2021 • Paper 1

Boiling Point Elevation of Silver Nitrate Solution

Comprehension Passage

The boiling point of water in a 0.1 molal0.1\text{ molal} silver nitrate solution (solution A) is x ∘Cx\text{ }^\circ\text{C}. To this solution A, an equal volume of 0.1 molal0.1\text{ 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×10−2 ∘Cy \times 10^{-2}\text{ }^\circ\text{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−1K_b = 0.5\text{ K kg mol}^{-1}; Boiling point of pure water as 100 ∘C100\text{ }^\circ\text{C}.)

The value of xx is ___.

Official Numerical Answer100.1

Topics & Concepts

SolutionsConcentration TermsMole Concept

Step-by-Step Solution

  1. Dissociation of Silver Nitrate in Solution A:
    AgNO3\text{AgNO}_3 is a strong electrolyte and dissociates completely in water:
    AgNO3(aq)⟶Ag+(aq)+NO3−(aq)\text{AgNO}_3(aq) \longrightarrow \text{Ag}^+(aq) + \text{NO}_3^-(aq)
    The van 't Hoff factor is i=2i = 2.

  2. Elevation in Boiling Point (ΔTb\Delta T_b):
    ΔTb,A=i⋅Kb⋅m\Delta T_{b,\text{A}} = i \cdot K_b \cdot m
    Given m=0.1 mol kg−1m = 0.1\text{ mol kg}^{-1} and Kb=0.5 K kg mol−1K_b = 0.5\text{ K kg mol}^{-1}:
    ΔTb,A=2×0.5×0.1=0.1 ∘C\Delta T_{b,\text{A}} = 2 \times 0.5 \times 0.1 = 0.1\ ^\circ\text{C}

  3. Boiling Point of Solution A (xx):
    x=Tb∘+ΔTb,A=100 ∘C+0.1 ∘C=100.1 ∘Cx = T_b^\circ + \Delta T_{b,\text{A}} = 100\ ^\circ\text{C} + 0.1\ ^\circ\text{C} = 100.1\ ^\circ\text{C}

Therefore, the value of xx is 100.1.

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