At `18^(@)C` the mobilities of `NH_(4)^(+)` and `CIO_(4)^(-)` ions are `6.6 xx 10^(-4)` and `5.7 xx 10^(-4)cm^(2)"volt"^(-1)sec^(-1)` at infinite dilution. Calculate equivalent conductance of ammonium chlorate solution.

Explanation

To calculate the equivalent conductance of ammonium chlorate (NH₄ClO₄) solution at 18°C, we can use the concept of ionic mobilities and Kohlrausch's law. Here’s a step-by-step solution:

### Step 1: Understand the given data
- The mobility of the ammonium ion (NH₄⁺) is given as:
\[ \mu_{NH_4^+} = 6.6 \times 10^{-4} \, \text{cm}^2 \, \text{V}^{-1} \, \text{s}^{-1} \]
- The mobility of the chlorate ion (ClO₄⁻) is given as:
\[ \mu_{ClO_4^-} = 5.7 \times 10^{-4} \, \text{cm}^2 \, \text{V}^{-1} \, \text{s}^{-1} \]

### Step 2: Use Kohlrausch's law
According to Kohlrausch's law, the equivalent conductance (Λₘ⁰) at infinite dilution for the salt (NH₄ClO₄) can be calculated using the sum of the contributions from its constituent ions:
\[ \Lambda_m^0(NH_4ClO_4) = \Lambda_m^0(NH_4^+) + \Lambda_m^0(ClO_4^-) \]

### Step 3: Relate mobility to conductance
The equivalent conductance can also be expressed in terms of the mobilities of the ions:
\[ \Lambda_m^0 = F \cdot (\mu_{NH_4^+} + \mu_{ClO_4^-}) \]
where \( F \) is Faraday's constant, approximately \( 96500 \, \text{C/mol} \).

### Step 4: Substitute the values
Now, substituting the values of mobilities and Faraday's constant:
\[ \Lambda_m^0(NH_4ClO_4) = 96500 \cdot (6.6 \times 10^{-4} + 5.7 \times 10^{-4}) \]

### Step 5: Calculate the sum of mobilities
Calculate the sum of the mobilities:
\[ \mu_{NH_4^+} + \mu_{ClO_4^-} = 6.6 \times 10^{-4} + 5.7 \times 10^{-4} = 12.3 \times 10^{-4} \, \text{cm}^2 \, \text{V}^{-1} \, \text{s}^{-1} \]

### Step 6: Calculate the equivalent conductance
Now, substitute this back into the equation:
\[ \Lambda_m^0(NH_4ClO_4) = 96500 \cdot (12.3 \times 10^{-4}) \]
\[ \Lambda_m^0(NH_4ClO_4) = 96500 \cdot 0.00123 = 118.67 \, \text{S cm}^2/\text{mol} \]

### Final Answer
The equivalent conductance of ammonium chlorate solution at 18°C is:
\[ \Lambda_m^0(NH_4ClO_4) \approx 118.67 \, \text{S cm}^2/\text{mol} \]