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The equivalent conductance of monobasic ...

The equivalent conductance of monobasic acid at infinite dilution is `438 ohm^(-1) cm^(2) eq^(-1)`. If the resistivity of the solution containing 15g of acid (Molecular weight =49) in 1L is 18.5 ohm cm. What is the degree of dissociation of acid.

A

0.459

B

0.402

C

0.604

D

0.507

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The correct Answer is:
To find the degree of dissociation of the monobasic acid, we can follow these steps: ### Step 1: Calculate the Normality of the Acid Solution Given: - Mass of the acid = 15 g - Molecular weight of the acid = 49 g/mol First, we calculate the number of equivalents of the acid: \[ \text{Number of equivalents} = \frac{\text{Mass of acid}}{\text{Molecular weight}} = \frac{15 \, \text{g}}{49 \, \text{g/mol}} \approx 0.306 \, \text{eq} \] Since the solution is 1 L, the normality (N) of the solution is: \[ N = \text{Number of equivalents} = 0.306 \, \text{N} \] ### Step 2: Calculate the Conductivity of the Solution Given: - Resistivity of the solution = 18.5 ohm cm The conductivity (\( \kappa \)) is the inverse of resistivity: \[ \kappa = \frac{1}{\text{Resistivity}} = \frac{1}{18.5 \, \text{ohm cm}} \approx 0.05405 \, \text{S/cm} \] ### Step 3: Calculate the Equivalent Conductance of the Solution The equivalent conductance (\( \Lambda \)) can be calculated using the formula: \[ \Lambda = \frac{\kappa}{N} \] Substituting the values: \[ \Lambda = \frac{0.05405 \, \text{S/cm}}{0.306 \, \text{N}} \approx 0.176 \, \text{S cm}^2/\text{eq} \] ### Step 4: Calculate the Degree of Dissociation The degree of dissociation (\( \alpha \)) can be calculated using the formula: \[ \alpha = \frac{\Lambda}{\Lambda_0} \] Where \( \Lambda_0 \) is the equivalent conductance at infinite dilution, given as 438 ohm\(^{-1}\) cm\(^{2}\) eq\(^{-1}\). Substituting the values: \[ \alpha = \frac{0.176 \, \text{S cm}^2/\text{eq}}{438 \, \text{S cm}^2/\text{eq}} \approx 0.0004 \] To express this as a percentage: \[ \alpha \times 100 \approx 0.04\% \] ### Step 5: Conclusion The degree of dissociation of the monobasic acid is approximately 0.04%.

To find the degree of dissociation of the monobasic acid, we can follow these steps: ### Step 1: Calculate the Normality of the Acid Solution Given: - Mass of the acid = 15 g - Molecular weight of the acid = 49 g/mol First, we calculate the number of equivalents of the acid: ...
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