The molar conductance of NaCl, HCl and at infinite dilution are 126.45, 426.16 and 91.0 S respectively. The molar conductance of at infinite dilution is. Choose the right option for your answer. [2021]

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Q.
The molar conductance of NaCl, HCl and ${CH}_{3} COONa$ at infinite dilution are 126.45, 426.16 and 91.0 S ${cm}^{2}$ ${mol}^{- 1}$ respectively. The molar conductance of ${CH}_{3} COOH$ at infinite dilution is. Choose the right option for your answer. [2021]

1 540.48 S ${cm}^{2}$ ${mol}^{- 1}$

2 201.28 S ${cm}^{2}$ ${mol}^{- 1}$

3 390.71 S ${cm}^{2}$ ${mol}^{- 1}$

4 698.28 S ${cm}^{2}$ ${mol}^{- 1}$

Ans.
(3)

$Λ_{{CH}_{3} COOH}^{\infty} = Λ_{{CH}_{3} COONa}^{\circ} + Λ_{HCl}^{\circ} - Λ_{NaCl}^{\circ}$

$= 91.0 + 426.16 - 126.45$

$= 390.71$ $S$ ${cm}^{2}$ ${mol}^{- 1}$

Solution

Q.
The molar conductance of NaCl, HCl and ${CH}_{3} COONa$ at infinite dilution are 126.45, 426.16 and 91.0 S ${cm}^{2}$ ${mol}^{- 1}$ respectively. The molar conductance of ${CH}_{3} COOH$ at infinite dilution is. Choose the right option for your answer. [2021]

1 540.48 S ${cm}^{2}$ ${mol}^{- 1}$

2 201.28 S ${cm}^{2}$ ${mol}^{- 1}$

3 390.71 S ${cm}^{2}$ ${mol}^{- 1}$

4 698.28 S ${cm}^{2}$ ${mol}^{- 1}$

Ans.
(3)

$Λ_{{CH}_{3} COOH}^{\infty} = Λ_{{CH}_{3} COONa}^{\circ} + Λ_{HCl}^{\circ} - Λ_{NaCl}^{\circ}$

$= 91.0 + 426.16 - 126.45$

$= 390.71$ $S$ ${cm}^{2}$ ${mol}^{- 1}$

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Solution

Q. The molar conductance of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm2 mol-1 respectively. The molar conductance of CH3COOH at infinite dilution is. Choose the right option for your answer. [2021]

1 540.48 S cm2 mol-1

2 201.28 S cm2 mol-1

3 390.71 S cm2 mol-1

4 698.28 S cm2 mol-1

Ans. (3) ΛCH3COOH∞ =ΛCH3COONa∘+ΛHCl∘-ΛNaCl∘ =91.0+426.16-126.45 =390.71 S cm2 mol-1

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Q.
The molar conductance of NaCl, HCl and ${CH}_{3} COONa$ at infinite dilution are 126.45, 426.16 and 91.0 S ${cm}^{2}$ ${mol}^{- 1}$ respectively. The molar conductance of ${CH}_{3} COOH$ at infinite dilution is. Choose the right option for your answer. [2021]

1 540.48 S ${cm}^{2}$ ${mol}^{- 1}$

2 201.28 S ${cm}^{2}$ ${mol}^{- 1}$

3 390.71 S ${cm}^{2}$ ${mol}^{- 1}$

4 698.28 S ${cm}^{2}$ ${mol}^{- 1}$

Ans.
(3)

$Λ_{{CH}_{3} COOH}^{\infty} = Λ_{{CH}_{3} COONa}^{\circ} + Λ_{HCl}^{\circ} - Λ_{NaCl}^{\circ}$

$= 91.0 + 426.16 - 126.45$

$= 390.71$ $S$ ${cm}^{2}$ ${mol}^{- 1}$