The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of \(1~\mathrm{cm}^2\). The conductance of this solution was found to be \(5 \times 10^{-7}~\mathrm{S}\). The pH of the solution is 4. The value of limiting molar conductivity \((\Lambda_m^o)\) of this weak monobasic acid in aqueous solution is \(Z \times 10^2~\mathrm{S~cm}^{-1}~\mathrm{mol}^{-1}\). The value of \(Z\) is

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Redox Reactions and Electrochemistry

The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of

1 cm^{2}

. The conductance of this solution was found to be

5 \times 1 0^{- 7} S

. The pH of the solution is 4. The value of limiting molar conductivity

(Λ_{m}^{o})

of this weak monobasic acid in aqueous solution is

Z \times 1 0^{2} S cm^{- 1} mol^{- 1}

. The value of

Z

Answer: 6

Solution

Cell constant

= \frac{ℓ}{a} = \frac{120}{1} = 120 cm^{- 1}

Conductivity:

κ = G \times \frac{ℓ}{a} = 5 \times 1 0^{- 7} \times 120 = 6 \times 1 0^{- 5} S cm^{- 1}

Molar conductivity at concentration C:

Λ_{m}^{C} = \frac{κ \times 1000}{C} = \frac{6 \times 1 0 ^{- 5} \times 1000}{0.0015} = 40 S cm^{2} mol^{- 1}

From pH = 4,

[H^{+}] = 1 0^{- 4} M

Degree of dissociation:

α = \frac{[ H ^{+} ]}{C} = \frac{1 0 ^{- 4}}{0.0015} = \frac{1}{15}

Also,

α = \frac{Λ _{m}^{C}}{Λ _{m}^{o}}

\frac{1}{15} = \frac{40}{Λ _{m}^{o}}

Λ_{m}^{o} = 600 S cm^{2} mol^{- 1} = 6 \times 1 0^{2} S cm^{2} mol^{- 1}

Therefore,

Z = 6

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer

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JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Redox Reactions and Electrochemistry

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