Find out the weight of `H_(2)SO_(4)` in `150 mL, (N)/(7)H_(2)SO_(4)`.

Options

A. `2.05` g
B. `1.05` g
C. `0.05` g
D. `1.5` g

Explanation

To find the weight of \( H_2SO_4 \) in 150 mL of \( \frac{1}{7} N \) \( H_2SO_4 \), we can follow these steps:

### Step 1: Understand the given data
We are given:
- Normality (N) = \( \frac{1}{7} \) N
- Volume (V) = 150 mL

### Step 2: Convert volume from mL to L
Since normality is expressed in terms of liters, we need to convert the volume from mL to L:
\[ V = 150 \, \text{mL} = \frac{150}{1000} \, \text{L} = 0.150 \, \text{L} \]

### Step 3: Determine the valency factor
For \( H_2SO_4 \), the basicity (valency factor) is 2 because it can donate 2 protons (H\(^+\)).

### Step 4: Use the formula for normality
Normality (N) is defined as:
\[ N = \frac{\text{number of equivalents}}{\text{volume in L}} \]
The number of equivalents can also be expressed as:
\[ \text{number of equivalents} = \frac{\text{weight}}{\text{equivalent weight}} \]
The equivalent weight of \( H_2SO_4 \) can be calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{valency factor}} = \frac{98 \, \text{g/mol}}{2} = 49 \, \text{g/equiv} \]

### Step 5: Set up the equation
Using the normality formula:
\[ \frac{1}{7} = \frac{\text{weight}}{49 \times 0.150} \]

### Step 6: Solve for weight
Rearranging the equation to find the weight:
\[ \text{weight} = \frac{1}{7} \times 49 \times 0.150 \]
Calculating this gives:
\[ \text{weight} = \frac{49 \times 0.150}{7} = \frac{7.35}{7} = 1.05 \, \text{g} \]

### Conclusion
The weight of \( H_2SO_4 \) in 150 mL of \( \frac{1}{7} N \) \( H_2SO_4 \) is **1.05 g**.

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