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 of NaNO3 at infinite dilution using the given molar conductances of AgNO3, AgCl, and NaCl, we will apply Kohlrausch's law, which states that the molar conductance of an electrolyte at infinite dilution is equal to the sum of the conductances of its constituent ions.

### Step-by-Step Solution:

1. **Identify the given values**:
- Molar conductance of AgNO3 (Λ°(AgNO3)) = 116.5 Ω^(-1) cm^2/mole
- Molar conductance of AgCl (Λ°(AgCl)) = 121.6 Ω^(-1) cm^2/mole
- Molar conductance of NaCl (Λ°(NaCl)) = 110.3 Ω^(-1) cm^2/mole

2. **Write the expression for NaNO3**:
- NaNO3 dissociates into Na⁺ and NO3⁻ ions. According to Kohlrausch's law:
\[ Λ°(NaNO3) = Λ°(Na⁺) + Λ°(NO3⁻) \]

3. **Set up the equation using the known values**:
- From the dissociation of AgNO3 and AgCl, we can express the conductance of NaNO3 in terms of the other given conductances:
\[ Λ°(NaNO3) = Λ°(NaCl) + Λ°(AgNO3) - Λ°(AgCl) \]

4. **Substitute the known values**:
\[ Λ°(NaNO3) = 110.3 + 116.5 - 121.6 \]

5. **Perform the calculations**:
- First, add the conductances of NaCl and AgNO3:
\[ 110.3 + 116.5 = 226.8 \]
- Then subtract the conductance of AgCl:
\[ 226.8 - 121.6 = 105.2 \]

6. **Final answer**:
- The molar conductance of NaNO3 at infinite dilution is:
\[ Λ°(NaNO3) = 105.2 Ω^(-1) cm^2/mole \]

### Conclusion:
The molar conductance of NaNO3 at infinite dilution is **105.2 Ω^(-1) cm^2/mole**.