Two bottles A and B contain `2`M and 2m aqueous solution of sulphuric acid respectively , then

Options

A. A is more concentrated than B
B. B is more concentrated than A
C. Conc. Of A and B are equal.
D. it is important to compare the concentrations

Explanation

To solve the problem, we need to compare the concentrations of the two solutions: one is a 2M (molar) solution of sulfuric acid (H₂SO₄) in bottle A, and the other is a 2m (molal) solution of sulfuric acid in bottle B.

### Step-by-Step Solution:

**Step 1: Understand Molarity and Molality**
- Molarity (M) is defined as the number of moles of solute per liter of solution.
- Molality (m) is defined as the number of moles of solute per kilogram of solvent.

**Step 2: Convert Molality to Molarity**
- To compare the two solutions, we need to convert the molality of the solution in bottle B to molarity.
- The formula to convert molality (m) to molarity (M) is:

\[ M = \frac{m \times D}{1 - m \times M_{solute}} \]

where:
- \( m \) is the molality,
- \( D \) is the density of the solution,
- \( M_{solute} \) is the molar mass of the solute.

**Step 3: Calculate Molarity for Bottle B**
- For sulfuric acid (H₂SO₄), the molar mass (M₁) is approximately 98 g/mol.
- Assume the density of the 2m solution is approximately 1 g/mL (or 1000 g/L) for simplicity.
- Using the formula:

\[ M = \frac{2 \, \text{mol/kg} \times 1000 \, \text{g/L}}{1 - 2 \, \text{mol/kg} \times 0.098 \, \text{kg/mol}} \]

- First, calculate \( 2 \times 0.098 = 0.196 \).
- Now, \( 1 - 0.196 = 0.804 \).
- Therefore,

\[ M = \frac{2000}{0.804} \approx 248.76 \, \text{M} \]

**Step 4: Compare the Molarities**
- Bottle A has a concentration of 2M.
- Bottle B has a calculated molarity of approximately 2.48M.
- Since 2.48M > 2M, the solution in bottle B is more concentrated than that in bottle A.

**Conclusion:**
- The solution in bottle B (2m) is more concentrated than the solution in bottle A (2M).