Specific conductance of 0.1M nitric acid is
`6.3xx10^(-2)ohm^(-1)cm^(-1)`. The molar conductance of the
solution is:

Options

A. `630ohm^(-1)cm^(2)"mol"^(-1)`
B. `315ohm^(-1)cm^(2)"mol"^(-1)`
C. `100ohm^(-1)cm^(2)"mol"^(-1)`
D. `6.300ohm^(-1)cm^(2)"mol"^(-1)`

Explanation

To find the molar conductance of a 0.1 M nitric acid solution given its specific conductance, we can follow these steps:

### Step 1: Understand the relationship between specific conductance and molar conductance
Molar conductance (Λ) can be calculated from specific conductance (κ) using the formula:
\[ \Lambda = \frac{\kappa \times 1000}{C} \]
where:
- \( \Lambda \) is the molar conductance in ohm\(^{-1}\)cm\(^2\)mol\(^{-1}\),
- \( \kappa \) is the specific conductance in ohm\(^{-1}\)cm\(^{-1}\),
- \( C \) is the concentration in mol/L.

### Step 2: Plug in the values
We are given:
- Specific conductance \( \kappa = 6.3 \times 10^{-2} \) ohm\(^{-1}\)cm\(^{-1}\)
- Concentration \( C = 0.1 \) mol/L

Now substituting these values into the formula:
\[ \Lambda = \frac{6.3 \times 10^{-2} \, \text{ohm}^{-1}\text{cm}^{-1} \times 1000 \, \text{cm}^3/\text{L}}{0.1 \, \text{mol/L}} \]

### Step 3: Calculate the numerator
Calculating the numerator:
\[ 6.3 \times 10^{-2} \times 1000 = 63 \, \text{ohm}^{-1}\text{cm}^{-1} \]

### Step 4: Divide by the concentration
Now, we divide by the concentration:
\[ \Lambda = \frac{63 \, \text{ohm}^{-1}\text{cm}^{-1}}{0.1 \, \text{mol/L}} = 630 \, \text{ohm}^{-1}\text{cm}^{2}\text{mol}^{-1} \]

### Step 5: Write the final answer
Thus, the molar conductance of the 0.1 M nitric acid solution is:
\[ \Lambda = 630 \, \text{ohm}^{-1}\text{cm}^{2}\text{mol}^{-1} \]