The molar conductanve at infinite dilution of `AgNO_(3),AgCl and NaCl ` are `116.5,121.6 and 110.3 Omega^(-1) cm^(2)" mole"^(-1)` respectively. The molar conductances of `NaNo_(3)` is:

Options

A. `111.4 Omega^(-1) cm^(2)"mole"^(-1)`
B. `105.2 Omega^(-1) cm^(2)"mole"^(-1)`
C. `130.6 Omega^(-1) cm^(2)"mole"^(-1)`
D. `150.2 Omega^(-1) cm^(2)"mole"^(-1)`

Answer

To find the molar conductance at infinite dilution of NaNO3, we can use the following relationship based on the dissociation of the compounds involved:

1. **Identify the dissociation of the compounds:**
- AgNO3 dissociates into Ag⁺ and NO3⁻.
- NaCl dissociates into Na⁺ and Cl⁻.
- AgCl dissociates into Ag⁺ and Cl⁻.

2. **Write the equation for molar conductance:**
The molar conductance at infinite dilution (λ) of NaNO3 can be calculated using the formula:
\[ \lambda_{\text{NaNO}_3} = \lambda_{\text{AgNO}_3} + \lambda_{\text{NaCl}} - \lambda_{\text{AgCl}} \]

3. **Substitute the given values:**
- λ(AgNO3) = 116.5 Ω⁻¹ cm² mole⁻¹
- λ(NaCl) = 110.3 Ω⁻¹ cm² mole⁻¹
- λ(AgCl) = 121.6 Ω⁻¹ cm² mole⁻¹

Plugging in these values:
\[ \lambda_{\text{NaNO}_3} = 116.5 + 110.3 - 121.6 \]

4. **Perform the calculation:**
\[ \lambda_{\text{NaNO}_3} = 116.5 + 110.3 - 121.6 = 105.2 \, \Omega^{-1} \, \text{cm}^2 \, \text{mole}^{-1} \]

5. **Conclusion:**
The molar conductance at infinite dilution of NaNO3 is 105.2 Ω⁻¹ cm² mole⁻¹.