Question

What is the normality of a solution containing 4.9 g of \( \text{H}_2\text{SO}_4 \) in 500 mL of solution? (Molar mass of \( \text{H}_2\text{SO}_4 \) = 98 g/mol)

• A. 0.1 N
• B. 0.2 N
• C. 0.5 N
• D. 1 N

Hint:

For acids: \[ \text{Normality} = \text{Molarity} \times \text{n-factor} \] Here, \( \text{H}_2\text{SO}_4 \) has n-factor = 2 since it can donate 2 protons.

Answer 1

The Correct Option is A

Step 1: Understand normality for acids. 
Normality (N) is defined as: \[ N = \frac{\text{Equivalents of solute}}{\text{Volume of solution in L}} \] Step 2: Find equivalents of \( \text{H}_2\text{SO}_4 \). 
Molar mass of \( \text{H}_2\text{SO}_4 \) = 98 g/mol 
Since it's a diprotic acid (2 replaceable H\(^+\)), its n-factor = 2 
\[ \text{Equivalents} = \frac{\text{Given mass} \times \text{n-factor}}{\text{Molar mass}} = \frac{4.9 \times 2}{98} = \frac{9.8}{98} = 0.1 \] Step 3: Convert volume to liters. 
\[ 500\, \text{mL} = 0.5\, \text{L} \] Step 4: Calculate Normality. 
\[ N = \frac{0.1}{0.5} = 0.2 \] Wait—this seems inconsistent. Let's double-check: \[ \text{Equivalents} = \frac{4.9 \times 2}{98} = \frac{9.8}{98} = 0.1 \] So we made an error above. Actually, \[ \frac{4.9 \times 2}{98} = \frac{9.8}{98} = 0.1 \text{ equivalents} \] \[ N = \frac{0.1}{0.5} = 0.2 \] So final answer is: Correct Answer: (2) 0.2 N