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Electrochemistry question

2021 · 18 Mar · Shift 2 · Q18
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  5. /2021 · 18 Mar · Shift 2 · Q18

Electrochemistry question

2021 · 18 Mar · Shift 2 · Q18

JEE MainChemistryElectrochemistryNumerical+4 / −1
The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm2 mol −-− 1 respectively. The molar conductivity at infinite dilution of barium sulphate is ‾\underline{\hspace{2cm}}​ S cm2 mol −-− 1. (Round off to the Nearest Integer ).
Numerical answer
View written solutionFree

Correct answer: 288

  1. Use Kohlrausch’s law of independent migration of ions

    At infinite dilution, the molar conductivity of an electrolyte is the sum of the ionic conductivities of its ions.

    Let: Λm∞(BaCl2)=λBa2+∞+2λCl−∞=280\Lambda_m^\infty(\text{BaCl}_2)=\lambda^\infty_{\text{Ba}^{2+}}+2\lambda^\infty_{\text{Cl}^-}=280Λm∞​(BaCl2​)=λBa2+∞​+2λCl−∞​=280 Λm∞(H2SO4)=2λH+∞+λSO42−∞=860\Lambda_m^\infty(\text{H}_2\text{SO}_4)=2\lambda^\infty_{\text{H}^+}+\lambda^\infty_{\text{SO}_4^{2-}}=860Λm∞​(H2​SO4​)=2λH+∞​+λSO42−​∞​=860 Λm∞(HCl)=λH+∞+λCl−∞=426\Lambda_m^\infty(\text{HCl})=\lambda^\infty_{\text{H}^+}+\lambda^\infty_{\text{Cl}^-}=426Λm∞​(HCl)=λH+∞​+λCl−∞​=426

  2. Write the required expression

    For barium sulphate: Λm∞(BaSO4)=λBa2+∞+λSO42−∞\Lambda_m^\infty(\text{BaSO}_4)=\lambda^\infty_{\text{Ba}^{2+}}+\lambda^\infty_{\text{SO}_4^{2-}}Λm∞​(BaSO4​)=λBa2+∞​+λSO42−​∞​

  3. Eliminate the common ions using the given data

    Add the first two equations:

    =(\lambda^\infty_{\text{Ba}^{2+}}+2\lambda^\infty_{\text{Cl}^-})+(2\lambda^\infty_{\text{H}^+}+\lambda^\infty_{\text{SO}_4^{2-}})$$ $$=\lambda^\infty_{\text{Ba}^{2+}}+\lambda^\infty_{\text{SO}_4^{2-}}+2(\lambda^\infty_{\text{H}^+}+\lambda^\infty_{\text{Cl}^-})$$ Since: $$\lambda^\infty_{\text{H}^+}+\lambda^\infty_{\text{Cl}^-}=\Lambda_m^\infty(\text{HCl})=426$$ Therefore, $$\Lambda_m^\infty(\text{BaSO}_4)=\Lambda_m^\infty(\text{BaCl}_2)+\Lambda_m^\infty(\text{H}_2\text{SO}_4)-2\Lambda_m^\infty(\text{HCl})$$

  4. Substitute the values

    Λm∞(BaSO4)=280+860−2(426)\Lambda_m^\infty(\text{BaSO}_4)=280+860-2(426)Λm∞​(BaSO4​)=280+860−2(426) =1140−852=1140-852=1140−852 =288=288=288

  5. Final answer

    288 S cm2 mol−1\boxed{288\ \text{S cm}^2\text{ mol}^{-1}}288 S cm2 mol−1​

    Rounded to the nearest integer: 288.

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