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 of finding the molarity of the resulting mixture when 100 mL of 5 m H₂SO₄ is mixed with 100 mL of 8 m H₂SO₄, we will follow these steps:

### 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
- Density = 1 g/mL
- Mass = Density × Volume = 1 g/mL × 100 mL = 100 g
- Convert mass to kg: 100 g = 0.1 kg

2. **For the second solution (8 m H₂SO₄, density = 1.25 g/mL)**:
- Volume = 100 mL
- Density = 1.25 g/mL
- Mass = Density × Volume = 1.25 g/mL × 100 mL = 125 g
- Convert mass to kg: 125 g = 0.125 kg

### Step 2: Calculate the number of moles of H₂SO₄ in each solution.

1. **For the first solution**:
- Molality (m) = moles of solute / mass of solvent (in kg)
- Given molality = 5 m
- Let the number of moles of H₂SO₄ be x.
- 5 = x / 0.1
- x = 5 × 0.1 = 0.5 moles

2. **For the second solution**:
- Given molality = 8 m
- Let the number of moles of H₂SO₄ be y.
- 8 = y / 0.125
- y = 8 × 0.125 = 1 mole

### Step 3: Calculate the total number of moles of H₂SO₄ in the mixture.

- Total moles = moles from first solution + moles from second solution
- Total moles = 0.5 + 1 = 1.5 moles

### Step 4: Calculate the total volume of the resulting solution.

- Since there is no change in volume upon mixing, the total volume = 100 mL + 100 mL = 200 mL = 0.2 L

### Step 5: Calculate the molarity of the resulting mixture.

- Molarity (M) = moles of solute / volume of solution (in L)
- M = 1.5 moles / 0.2 L = 7.5 M

### Final Answer:
The molarity of the resulting mixture is **7.5 M**.

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