The molar conductivities of KCl, NaCl and `KNO_(3)` are 152,128 and `111" S cm"^(2) mol^(-1)` respectively. What is the molar conductivity of `NaNO_(3)` ?

Options

A. `101 S cm^(2) mol^(-1)`
B. `87 S cm^(2) mol^(-1)`
C. `-101 S cm^(2) mol^(-1)`
D. `-391S cm^(2) mol^(-1)`

Answer

To find the molar conductivity of NaNO₃, we can use the relationship between the molar conductivities of the given salts. The molar conductivity of a salt can be determined using the following formula:

\[ \Lambda_{m}(\text{NaNO}_3) = \Lambda_{m}(\text{NaCl}) + \Lambda_{m}(\text{KNO}_3) - \Lambda_{m}(\text{KCl}) \]

Where:
- \(\Lambda_{m}(\text{NaNO}_3)\) is the molar conductivity of NaNO₃.
- \(\Lambda_{m}(\text{NaCl})\) is the molar conductivity of NaCl.
- \(\Lambda_{m}(\text{KNO}_3)\) is the molar conductivity of KNO₃.
- \(\Lambda_{m}(\text{KCl})\) is the molar conductivity of KCl.

**Step 1: Write down the known values.**
- \(\Lambda_{m}(\text{KCl}) = 152 \, \text{S cm}^{-2} \text{mol}^{-1}\)
- \(\Lambda_{m}(\text{NaCl}) = 128 \, \text{S cm}^{-2} \text{mol}^{-1}\)
- \(\Lambda_{m}(\text{KNO}_3) = 111 \, \text{S cm}^{-2} \text{mol}^{-1}\)

**Step 2: Substitute the known values into the formula.**
\[ \Lambda_{m}(\text{NaNO}_3) = 128 + 111 - 152 \]

**Step 3: Perform the arithmetic.**
\[ \Lambda_{m}(\text{NaNO}_3) = 239 - 152 = 87 \, \text{S cm}^{-2} \text{mol}^{-1} \]

**Final Answer:**
The molar conductivity of NaNO₃ is \(87 \, \text{S cm}^{-2} \text{mol}^{-1}\).

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