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CBSE Class 12 Chemistry 2017 Delhi Set 1 Solved Paper

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Question : 9 of 26
Marks: +1, -0
Calculate the degree of dissociation (α)(\alpha) of acetic acid if its molar conductivity (Λm)(\Lambda_m) is 39.05 S cm39.05\ \mathrm{S}\ \mathrm{cm}  mol−1\ \mathrm{mol}^{-1}.
Given λ∘(H+)=349.6 S cm2 mol−1\lambda^{\circ} (\mathrm{H}^{+})=349.6\ \mathrm{S}\ \mathrm{cm}^2\ \mathrm{mol}^{-1} and λ∘(CH3COO−)\lambda^{\circ} (\mathrm{CH}_3\mathrm{COO}^{-}) =40.9 S cm2 mol−1=40.9\ \mathrm{S}\ \mathrm{cm}^2\ \mathrm{mol}^{-1}.
Solution:  
ΛCH3COOH∘=λCH3COO−∘+λH+∘\Lambda^{\circ}_{\mathrm{CH}_3\mathrm{COOH}} = \lambda^{\circ}_{\mathrm{CH}_3\mathrm{COO}^{-}} + \lambda^{\circ}_{\mathrm{H}^{+}}
  =40.9+349.6\;=40.9+349.6
  =390.5 S cm2/mol\;=390.5\ \mathrm{S}\ \mathrm{cm}^2 / \mathrm{mol}
N,    α  =ΛmΛm∘N,\;\;\alpha\;=\frac{\Lambda_m}{\Lambda^{\circ}_m}
  =39.05390.5=0.1\;=\frac{39.05}{390.5}=0.1
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