What is the molarity of `H_2SO_4` solution which contains 98% `H_2SO_4` by weight and whose specific gravity is 1.84 ?

Options

A. 1.84
B. 18.4
C. 9.2
D. 9.6

Answer

To find the molarity of the `H2SO4` solution that contains 98% `H2SO4` by weight and has a specific gravity of 1.84, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the given values:**
- Percentage by weight (W/W) of `H2SO4` = 98%
- Specific gravity (SG) = 1.84

2. **Convert specific gravity to density:**
- Since specific gravity is the ratio of the density of a substance to the density of water, we can calculate the density (D) of the solution:
\[ D = SG \times \text{Density of water} = 1.84 \times 1 \, \text{g/cm}^3 = 1.84 \, \text{g/cm}^3 \]

3. **Use the formula for molarity:**
- The formula for molarity (M) in terms of weight percentage is:
\[ M = \frac{x \times D \times 10}{M_a} \]
Where:
- \( x \) = weight percentage of solute = 98
- \( D \) = density of the solution = 1.84 g/cm³
- \( M_a \) = molar mass of `H2SO4`

4. **Calculate the molar mass of `H2SO4`:**
- The molar mass (M_a) of `H2SO4` can be calculated as follows:
\[ M_a = 2 \times 1 + 32 + 64 = 98 \, \text{g/mol} \]

5. **Substitute the values into the molarity formula:**
\[ M = \frac{98 \times 1.84 \times 10}{98} \]

6. **Simplify the equation:**
- The 98 in the numerator and denominator cancels out:
\[ M = 1.84 \times 10 = 18.4 \, \text{M} \]

7. **Final Answer:**
- The molarity of the `H2SO4` solution is **18.4 M**.