100 ml of 5 m `H_2SO_4` of density 1 gm/ml is mixed with 100 ml of 8 m `H_2SO_4` of density 1.25 g/mL. If there is no change in volume of resulting solution due to mixing, the molarity of the resulting mixture is -

Options

A. 5.5 M
B. 6.5 M
C. 7.5 M
D. 5.26 M

Answer

To solve the problem, we need to calculate the molarity of the resulting mixture after mixing two solutions of sulfuric acid (H₂SO₄). Here are the steps to find the solution:

### Step 1: Calculate the mass of H₂SO₄ in each solution
1. **For the first solution (5 m H₂SO₄, density = 1 g/mL):**
- Volume = 100 mL = 0.1 L
- Molarity (M) = 5 m
- Moles of H₂SO₄ = Molarity × Volume = 5 moles/L × 0.1 L = 0.5 moles
- Mass of H₂SO₄ = Moles × Molar Mass (H₂SO₄ = 98 g/mol) = 0.5 moles × 98 g/mol = 49 g

2. **For the second solution (8 m H₂SO₄, density = 1.25 g/mL):**
- Volume = 100 mL = 0.1 L
- Molarity (M) = 8 m
- Moles of H₂SO₄ = Molarity × Volume = 8 moles/L × 0.1 L = 0.8 moles
- Mass of H₂SO₄ = Moles × Molar Mass = 0.8 moles × 98 g/mol = 78.4 g

### Step 2: Calculate the total moles of H₂SO₄ in the mixture
- Total moles of H₂SO₄ = Moles from first solution + Moles from second solution
- Total moles = 0.5 moles + 0.8 moles = 1.3 moles

### Step 3: Calculate the total volume of the resulting solution
- Since it is given that there is no change in volume upon mixing, the total volume remains:
- Total volume = 100 mL + 100 mL = 200 mL = 0.2 L

### Step 4: Calculate the molarity of the resulting mixture
- Molarity (M) = Total moles of solute / Total volume of solution in liters
- Molarity = 1.3 moles / 0.2 L = 6.5 M

### Final Answer
The molarity of the resulting mixture is **6.5 M**.

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