What is the molarity of `H_(2)SO_(4)` solution that has a density 1.84 g/c c at `35^(@)`C and contains 98% by weight?

Options

A. (a)4.18 M
B. (b)8.14 M
C. (c )18.4 M
D. (d)18 M

Explanation

To find the molarity of the `H₂SO₄` solution, we will follow these steps:

### Step 1: Determine the mass of `H₂SO₄` in the solution
Given that the solution contains 98% by weight of `H₂SO₄`, in 100 grams of the solution, the mass of `H₂SO₄` can be calculated as:
\[ \text{Mass of } H₂SO₄ = 98\% \text{ of } 100 \text{ g} = 98 \text{ g} \]

### Step 2: Calculate the volume of the solution
The density of the solution is given as 1.84 g/cm³. We can use the formula for density:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Rearranging this gives us:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the values:
\[ \text{Volume} = \frac{100 \text{ g}}{1.84 \text{ g/cm}^3} \approx 54.35 \text{ cm}^3 \]
Since 1 cm³ = 1 mL, the volume in liters is:
\[ \text{Volume in liters} = \frac{54.35 \text{ mL}}{1000} = 0.05435 \text{ L} \]

### Step 3: Calculate the number of moles of `H₂SO₄`
The molar mass of `H₂SO₄` can be calculated as follows:
\[ \text{Molar mass of } H₂SO₄ = (2 \times 1) + 32 + (4 \times 16) = 2 + 32 + 64 = 98 \text{ g/mol} \]
Now, we can calculate the number of moles of `H₂SO₄`:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{98 \text{ g}}{98 \text{ g/mol}} = 1 \text{ mol} \]

### Step 4: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}} = \frac{1 \text{ mol}}{0.05435 \text{ L}} \approx 18.4 \text{ M} \]

Thus, the molarity of the `H₂SO₄` solution is approximately **18.4 M**.