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Conductivity of 2.5 x 10^-4 M methanoic acid is 5.25 x 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation.

User Mkocabas
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Final answer:

The molar conductivity of methanoic acid is calculated using the formula λ = K/c, resulting in 210 S cm^2 mol^-1, while the degree of dissociation requires the molar conductivity at infinite dilution, which was not provided in the data.

Step-by-step explanation:

To calculate the molar conductivity (λ) of methanoic acid, we use the formula λ = K/c, where K is the conductivity and c is the concentration. Given that K = 5.25 x 10-5 S cm-1 and c = 2.5 x 10-4 M, the molar conductivity is λ = (5.25 x 10-5 S cm-1) / (2.5 x 10-4 M) = 210 S cm2 mol-1.

To calculate the degree of dissociation (α), we can use the formula α = λ / λ₀, where λ₀ is the molar conductivity at infinite dilution for the electrolyte. Since we do not have λ₀ provided, we'll assume this question is theoretical or that λ₀ value is available elsewhere. If λ₀ is given, the degree of dissociation α can be found by plugging the values into the equation.

User Kevin Pang
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