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ChemHardAdv 2021 • Paper 1

Difference in Boiling Points after Precipitation Reaction

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 ∣y∣|y| is ___.

Official Numerical Answer2.5

Topics & Concepts

SolutionsConcentration TermsStoichiometry

Step-by-Step Solution

  1. Precipitation Reaction between Solution A and BaCl2\text{BaCl}_2:
    Let equal volumes VV of 0.1 molal0.1\text{ molal} AgNO3\text{AgNO}_3 and 0.1 molal0.1\text{ molal} BaCl2\text{BaCl}_2 be mixed.
    Since densities equal that of water (1 kg L−11\text{ kg L}^{-1}), 1 L1\text{ L} corresponds to 1 kg1\text{ kg} of solvent:
  • Initial moles of AgNO3=0.1×V\text{AgNO}_3 = 0.1 \times V
  • Initial moles of BaCl2=0.1×V\text{BaCl}_2 = 0.1 \times V

The precipitation reaction is:
2AgNO3(aq)+BaCl2(aq)⟶2AgCl(s)↓+Ba(NO3)2(aq)2\text{AgNO}_3(aq) + \text{BaCl}_2(aq) \longrightarrow 2\text{AgCl}(s)\downarrow + \text{Ba(NO}_3)_2(aq)

AgNO3\text{AgNO}_3 is the limiting reagent:

  • AgNO3\text{AgNO}_3 reacted =0.1V mol  ⟹  remaining =0= 0.1V\text{ mol} \implies \text{remaining } = 0
  • BaCl2\text{BaCl}_2 reacted =0.1V2=0.05V mol  ⟹  remaining =0.05V mol= \frac{0.1V}{2} = 0.05V\text{ mol} \implies \text{remaining } = 0.05V\text{ mol}
  • Ba(NO3)2\text{Ba(NO}_3)_2 formed =0.05V mol= 0.05V\text{ mol}
  • Precipitated AgCl(s)\text{AgCl}(s) is insoluble and does not contribute to boiling point elevation.
  1. Total Dissociated Ions in Solution B:
  • Remaining BaCl2\text{BaCl}_2 dissociates into 1 Ba2++2 Cl−1\text{ Ba}^{2+} + 2\text{ Cl}^- (33 ions):
    Moles of ions=3×0.05V=0.15V mol\text{Moles of ions} = 3 \times 0.05V = 0.15V\text{ mol}
  • Formed Ba(NO3)2\text{Ba(NO}_3)_2 dissociates into 1 Ba2++2 NO3−1\text{ Ba}^{2+} + 2\text{ NO}_3^- (33 ions):
    Moles of ions=3×0.05V=0.15V mol\text{Moles of ions} = 3 \times 0.05V = 0.15V\text{ mol}
    Total moles of ions in Solution B=0.15V+0.15V=0.30V mol\text{Total moles of ions in Solution B} = 0.15V + 0.15V = 0.30V\text{ mol}

  1. Effective Molality and Elevation of Solution B:
    Total mass of solvent in Solution B =V+V=2V kg= V + V = 2V\text{ kg}.
    Effective ion molality meff=0.30V mol2V kg=0.15 mol kg−1\text{Effective ion molality } m_{\text{eff}} = \frac{0.30V\text{ mol}}{2V\text{ kg}} = 0.15\text{ mol kg}^{-1}
    ΔTb,B=Kb⋅meff=0.5×0.15=0.075 ∘C\Delta T_{b,\text{B}} = K_b \cdot m_{\text{eff}} = 0.5 \times 0.15 = 0.075\ ^\circ\text{C}

  2. Difference in Boiling Points (y×10−2 ∘Cy \times 10^{-2}\ ^\circ\text{C}):
    ΔT=ΔTb,A−ΔTb,B=0.100 ∘C−0.075 ∘C=0.025 ∘C=2.5×10−2 ∘C\Delta T = \Delta T_{b,\text{A}} - \Delta T_{b,\text{B}} = 0.100\ ^\circ\text{C} - 0.075\ ^\circ\text{C} = 0.025\ ^\circ\text{C} = 2.5 \times 10^{-2}\ ^\circ\text{C}

Comparing with y×10−2 ∘Cy \times 10^{-2}\ ^\circ\text{C}:
y=2.5  ⟹  ∣y∣=2.5y = 2.5 \implies |y| = 2.5

Therefore, the value of ∣y∣|y| is 2.5.

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