Molar conductance of 1M solution of weak acid HA is 20 `ohm^(-1) cm^(2) mol^(-1)`. Find % dissocation of HA:
`(^^_(m)^(o)(H^(+)) =350 S cm^(2) mol^(-1)),(^^_(m)^(o)(A^(-)) =50 S cm^(2) mol^(-1))`

Answer

To find the percentage dissociation of the weak acid HA, we can follow these steps:

### Step 1: Understand the given data
We have:
- Molar conductance of 1M solution of weak acid HA, \( \Lambda_m = 20 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)
- Molar conductance at infinite dilution for \( H^+ \), \( \Lambda_m^0(H^+) = 350 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)
- Molar conductance at infinite dilution for \( A^- \), \( \Lambda_m^0(A^-) = 50 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)

### Step 2: Calculate the molar conductance at infinite dilution for HA
The molar conductance at infinite dilution for the weak acid HA can be calculated as the sum of the molar conductances of its ions:
\[ \Lambda_m^0(HA) = \Lambda_m^0(H^+) + \Lambda_m^0(A^-) = 350 + 50 = 400 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]

### Step 3: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^0(HA)} \]
Substituting the values:
\[ \alpha = \frac{20}{400} = \frac{1}{20} \]

### Step 4: Calculate the percentage dissociation
To find the percentage dissociation, we multiply α by 100:
\[ \text{Percentage dissociation} = \alpha \times 100 = \frac{1}{20} \times 100 = 5\% \]

### Final Answer
The percentage dissociation of the weak acid HA is **5%**.

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