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 \( \frac{N}{7} H_2SO_4 \), we can follow these steps:

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

### Step 2: Calculate the equivalent weight of \( H_2SO_4 \)
The equivalent weight of an acid is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Basicity}} \]
- The molecular weight of \( H_2SO_4 \) is calculated as follows:
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Sulfur (S): 32 g/mol × 1 = 32 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol
- Total = \( 2 + 32 + 64 = 98 \) g/mol

- The basicity of \( H_2SO_4 \) is 2 (since it can donate 2 protons, \( H^+ \)).
- Therefore, the equivalent weight is:
\[ \text{Equivalent weight} = \frac{98 \text{ g/mol}}{2} = 49 \text{ g/equiv} \]

### Step 3: Use the normality formula to find the weight (W)
The formula relating normality, weight, and volume is:
\[ N = \frac{W \times 1000}{\text{Equivalent weight} \times V} \]
Rearranging this to find \( W \):
\[ W = N \times \text{Equivalent weight} \times \frac{V}{1000} \]

### Step 4: Substitute the values into the equation
- Substitute \( N = \frac{1}{7} \), \( \text{Equivalent weight} = 49 \) g/equiv, and \( V = 150 \) mL:
\[ W = \left(\frac{1}{7}\right) \times 49 \times \frac{150}{1000} \]

### Step 5: Calculate \( W \)
\[ W = \frac{49 \times 150}{7 \times 1000} \]
\[ W = \frac{7350}{7000} = 1.05 \text{ g} \]

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

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