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Question: A solution of 0.2 M $CH_3COOH$ is placed between parallel electrodes of cross-section area 4 $cm^2$,...

A solution of 0.2 M CH3COOHCH_3COOH is placed between parallel electrodes of cross-section area 4 cm2cm^2, separated by 2 cm. For this solution, resistance measured is 100Ω\Omega. Calculate elevation in boiling point of the 0.2 M CH3COOHCH_3COOH solution, using following information.

KbK_b = 0.5 K kg/mol; Λm∞(H+)=150Scm2mol−1\Lambda_m^\infty (H^+) = 150 S cm^2 mol^{-1}; Λm∞(CH3COO−)=50Scm2mol−1\Lambda_m^\infty (CH_3COO^-) = 50 S cm^2 mol^{-1}.

[Assume Molarity = Molality]

Answer

Elevation in boiling point ≈ 0.11 K

Explanation

Solution

Solution:

  1. Determine cell constant (K):

    K=lA=2 cm4 cm2=0.5 cm−1K = \frac{l}{A} = \frac{2\,\text{cm}}{4\,\text{cm}^2} = 0.5\,\text{cm}^{-1}

  2. Calculate conductivity (κ):

    κ=KR=0.5100=0.005 S/cm\kappa = \frac{K}{R} = \frac{0.5}{100} = 0.005\,\text{S/cm}

  3. Theoretical molar conductivity for full dissociation:

    Given limiting molar conductivities Λm(H+)=150 S cm2/mol\Lambda_m(H^+) = 150\,\text{S\,cm}^2/\text{mol} and Λm(CH3COO−)=50 S cm2/mol\Lambda_m(CH_3COO^-) = 50\,\text{S\,cm}^2/\text{mol}

    Λm0=150+50=200 S cm2/mol\Lambda_m^0 = 150 + 50 = 200\,\text{S\,cm}^2/\text{mol}

  4. Concentration in mol/cm³:

    c=0.2 mol1000 cm3=2.0×10−4 mol/cm3c = \frac{0.2\,\text{mol}}{1000\,\text{cm}^3} = 2.0 \times 10^{-4}\,\text{mol/cm}^3

  5. Degree of dissociation (α):

    The conductivity for full dissociation would be: κ0=Λm0 c=200×2.0×10−4=0.04 S/cm\kappa_0 = \Lambda_m^0 \, c = 200 \times 2.0 \times 10^{-4} = 0.04\,\text{S/cm}

    Since observed conductivity is 0.005 S/cm, α=κobsκ0=0.0050.04=0.125\alpha = \frac{\kappa_{\text{obs}}}{\kappa_0} = \frac{0.005}{0.04} = 0.125

  6. Effective van’t Hoff factor (i):

    Acetic acid dissociates as: CH3COOH⇌H++CH3COO−CH_3COOH \rightleftharpoons H^+ + CH_3COO^-

    For each mole, if fraction α\alpha dissociates:

    • Moles undissociated: 0.2(1−α)0.2(1-\alpha)
    • Moles of ions: 2×0.2α2 \times 0.2\alpha

    Total effective moles = 0.2(1−α)+0.4α=0.2(1+α)0.2(1-\alpha) + 0.4\alpha = 0.2(1 + \alpha)

    Thus, i=1+α=1+0.125=1.125i = 1 + \alpha = 1 + 0.125 = 1.125

  7. Boiling point elevation (ΔT₍b₎):

    Using the formula: ΔTb=i Kb m\Delta T_b = i\, K_b\, m

    where Kb=0.5 K\cdotpkg/molK_b = 0.5\,\text{K·kg/mol} and m=0.2 mol/kgm = 0.2\,\text{mol/kg}, ΔTb=1.125×0.5×0.2=0.1125 K\Delta T_b = 1.125 \times 0.5 \times 0.2 = 0.1125\,\text{K}

Core Explanation:

  • Compute cell constant and conductivity.
  • Use limiting molar conductivity to find α.
  • Effective van’t Hoff factor i = 1 + α.
  • ΔT_b = i K_b m.