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Try this 5.3.7

Q.Calculate Λ0\Lambda_0 (CH2_2ClOOH) if Λ0\Lambda_0 values for HCl, KCl and CH2_2ClCOOK are repectively, 4.261, 1.499 and 1.132 Ω−1\Omega^{-1} cm2^2 mol−1^{-1}.
[!NOTE]
The book prints the target acid's formula as "CH2_2ClOOH" (missing the C of the -COOH group) and "repectively" -- transcribed as printed. The context makes clear that chloroacetic acid, CH2_2ClCOOH, is meant (its potassium salt CH2_2ClCOOK is among the given data), and the solution uses the correct formula.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Step 1. By the same logic used for acetic acid (section 5.3.7): Λ0(HCl)=λH+0+λCl−0\Lambda^0(HCl)=\lambda^0_{H^+}+\lambda^0_{Cl^-}, Λ0(CH2ClCOOK)=λK+0+λCH2ClCOO−0\Lambda^0(CH_2ClCOOK)=\lambda^0_{K^+}+\lambda^0_{CH_2ClCOO^-}, Λ0(KCl)=λK+0+λCl−0\Lambda^0(KCl)=\lambda^0_{K^+}+\lambda^0_{Cl^-}.

Step 2. Adding the first two and subtracting the third cancels λK+0\lambda^0_{K^+} and λCl−0\lambda^0_{Cl^-}, leaving exactly λH+0+λCH2ClCOO−0=Λ0(CH2ClCOOH)\lambda^0_{H^+}+\lambda^0_{CH_2ClCOO^-}=\Lambda^0(CH_2ClCOOH).

Step 3. Substituting the given values: Λ0(CH2ClCOOH)=4.261+1.132−1.499=3.894 Ω−1cm2mol−1\Lambda^0(CH_2ClCOOH)=4.261+1.132-1.499=3.894\ \Omega^{-1}cm^2mol^{-1}. …

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