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Solutions question

2010 · Shift 0 · Q23
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Solutions question

2010 · Shift 0 · Q23

JEE MainChemistrySolutionsMCQ+4 / −1
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (∆Tf), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1)
  1. A
    0.0372 K
  2. B
    0.0558 K
  3. C
    0.0744 K
  4. D
    0.0186 K
View written solutionFree

Correct answer: B

  1. Use freezing point depression formula

    ΔTf=iKfm\Delta T_f = i K_f mΔTf​=iKf​m

    where:

    • iii = van’t Hoff factor
    • Kf=1.86 K kg mol−1K_f = 1.86\ \text{K kg mol}^{-1}Kf​=1.86 K kg mol−1
    • mmm = molality

  2. Find molality

    Given 0.010.010.01 mol of Na2SO4\text{Na}_2\text{SO}_4Na2​SO4​ dissolved in 111 kg water:

    m=0.011=0.01 mol kg−1m = \frac{0.01}{1} = 0.01\ \text{mol kg}^{-1}m=10.01​=0.01 mol kg−1

  3. Find van’t Hoff factor

    Sodium sulphate dissociates completely as:

    Na2SO4→2Na++SO42−\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}Na2​SO4​→2Na++SO42−​

    Total ions formed =3= 3=3, so

    i=3i = 3i=3

  4. Calculate depression in freezing point

    ΔTf=3×1.86×0.01\Delta T_f = 3 \times 1.86 \times 0.01ΔTf​=3×1.86×0.01

    ΔTf=0.0558 K\Delta T_f = 0.0558\ \text{K}ΔTf​=0.0558 K

  5. Match with options

    0.0558 K0.0558\ \text{K}0.0558 K corresponds to Option B.

Final Answer

Option B: 0.0558 K0.0558\ \text{K}0.0558 K

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