Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options given.
`{:("Column I","Column II"),("For a 5% solution of "H_(2)SO_(4)(d ="1.01 g mL"^(-1)),),("(A) Molarity of the solution","(p) 0.537"),("(B) Molality of the solution","(q) 0.0096"),("(C) Mole fraction of "H_(2)SO_(4),"(r) 0.05"),("(D) Mass fraction of "H_(2)SO_(4),"(s) 0.515"):}`

Options

A. A-r, B-s, C-p, D-q
B. A-q, B-p, C-s, D-r
C. A-s, B-p, C-q, D-r
D. A-s, B-r, C-p, D-q

Explanation

To solve the problem, we need to match the entries from Column I with the appropriate entries from Column II for a 5% solution of \( H_2SO_4 \) with a density of \( 1.01 \, \text{g/mL} \).

### Step-by-Step Solution:

1. **Understanding the 5% Solution**:
- A 5% solution of \( H_2SO_4 \) means there are 5 grams of \( H_2SO_4 \) in 100 grams of solution.

2. **Calculating the Mass of Solvent**:
- The mass of the solvent (water) in the solution is:
\[ \text{Mass of solvent} = 100 \, \text{g (solution)} - 5 \, \text{g (solute)} = 95 \, \text{g} \]

3. **Calculating the Volume of the Solution**:
- Using the density to find the volume of the solution:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \, \text{g}}{1.01 \, \text{g/mL}} \approx 99.01 \, \text{mL} = 0.099 \, \text{L} \]

4. **Calculating Molarity (M)**:
- Molarity is defined as:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
- First, calculate the moles of \( H_2SO_4 \):
\[ \text{Molar mass of } H_2SO_4 = 98 \, \text{g/mol} \]
\[ \text{Moles of } H_2SO_4 = \frac{5 \, \text{g}}{98 \, \text{g/mol}} \approx 0.051 \, \text{mol} \]
- Now, calculate the molarity:
\[ M = \frac{0.051 \, \text{mol}}{0.099 \, \text{L}} \approx 0.515 \, \text{M} \]

5. **Calculating Molality (m)**:
- Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.051 \, \text{mol}}{0.095 \, \text{kg}} \approx 0.537 \, \text{mol/kg} \]

6. **Calculating Mole Fraction (X)**:
- Mole fraction of \( H_2SO_4 \):
\[ \text{Total moles} = \text{moles of } H_2SO_4 + \text{moles of water} \]
- Moles of water (molar mass = 18 g/mol):
\[ \text{Moles of water} = \frac{95 \, \text{g}}{18 \, \text{g/mol}} \approx 5.28 \, \text{mol} \]
- Total moles:
\[ \text{Total moles} = 0.051 + 5.28 \approx 5.331 \, \text{mol} \]
- Mole fraction of \( H_2SO_4 \):
\[ X_{H_2SO_4} = \frac{0.051}{5.331} \approx 0.0096 \]

7. **Calculating Mass Fraction**:
- Mass fraction of \( H_2SO_4 \):
\[ \text{Mass fraction} = \frac{\text{mass of } H_2SO_4}{\text{total mass}} = \frac{5 \, \text{g}}{100 \, \text{g}} = 0.05 \]

### Final Matching:
- (A) Molarity of the solution → (s) 0.515
- (B) Molality of the solution → (p) 0.537
- (C) Mole fraction of \( H_2SO_4 \) → (q) 0.0096
- (D) Mass fraction of \( H_2SO_4 \) → (r) 0.05