Question

One l of 1 M H$_2$SO$_4$ contains ----g of H$_2$SO$_4$

• A. 98
• B. 9.8
• C. 49
• D. 4.9

Hint:

To prepare standard solutions:
- 1 M \(\text{H}_2\text{SO}_4\) requires dissolving 1 mole (98 g) of the acid in 1 Liter of water.
- 1 N \(\text{H}_2\text{SO}_4\) requires dissolving 1 equivalent weight (Molar mass acidity/basicity = 98 2 = 49 g) of the acid in 1 Liter of water.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Molarity (\(M\)) is a concentration term defined as the number of moles of solute dissolved per liter of solution:
\[ \text{Molarity } (M) = \frac{\text{moles of solute}}{\text{Volume of solution in Liters}} \]
Key Formula or Approach:
To find the mass of solute in a given volume and molarity:
\[ \text{Mass of solute } (\text{g}) = \text{Molarity } (M) \times \text{Volume } (\text{L}) \times \text{Molar Mass } (\text{g/mol}) \]

Step 2: Detailed Explanation:
Let us calculate the mass of sulfuric acid (\(\text{H}_2\text{SO}_4\)) required:
1. Find the molar mass of \(\text{H}_2\text{SO}_4\):
- Hydrogen (\(\text{H}\)): \( 2 \times 1.008 = 2.016 \, \text{g/mol} \)
- Sulfur (\(\text{S}\)): \( 1 \times 32.06 = 32.06 \, \text{g/mol} \)
- Oxygen (\(\text{O}\)): \( 4 \times 16.00 = 64.00 \, \text{g/mol} \)
\[ \text{Molar Mass of H}_2\text{SO}_4 = 2.016 + 32.06 + 64.00 \approx 98.08 \, \text{g/mol} \]
2. Substitute the values into the formula:
- Molarity (\(M\)) = \( 1 \, \text{M} \)
- Volume (\(\text{V}\)) = \( 1 \, \text{L} \)
\[ \text{Mass} = 1 \, \text{mol/L} \times 1 \, \text{L} \times 98 \, \text{g/mol} = 98 \, \text{g} \]
Therefore, 1 Liter of a 1 M \(\text{H}_2\text{SO}_4\) solution contains exactly 98 grams of sulfuric acid.

Step 3: Final Answer:
One liter of 1 M \(\text{H}_2\text{SO}_4\) contains 98 g of \(\text{H}_2\text{SO}_4\).