What is the molarityk and molality of a 13% solution (by weight) of sulphric acid with a density of `1.02 mL^(-1)`? To what volume should `100 mL` of this acid be diluted in order to preapre a `1.5 N` solution?

Explanation

To solve the problem of finding the molarity and molality of a 13% solution of sulfuric acid and determining the volume to which 100 mL of this acid should be diluted to prepare a 1.5 N solution, we will follow these steps:

### Step 1: Calculate the mass of sulfuric acid in the solution
Given that the solution is 13% by weight, if we assume we have 100 g of the solution:
- Mass of sulfuric acid (H₂SO₄) = 13% of 100 g = 13 g

**Hint:** To find the mass of the solute in a percentage solution, multiply the total mass of the solution by the percentage (as a decimal).

### Step 2: Calculate the number of moles of sulfuric acid
The molar mass of sulfuric acid (H₂SO₄) is approximately 98 g/mol.
- Number of moles of H₂SO₄ = mass / molar mass = 13 g / 98 g/mol = 0.1327 mol

**Hint:** To find the number of moles, divide the mass of the solute by its molar mass.

### Step 3: Calculate the volume of the solution
Given the density of the solution is 1.02 g/mL:
- Volume of the solution = mass / density = 100 g / 1.02 g/mL = 98.04 mL = 0.09804 L

**Hint:** To find the volume of a solution, divide the mass of the solution by its density.

### Step 4: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution.
- Molarity = moles of solute / volume of solution in liters = 0.1327 mol / 0.09804 L = 1.352 M

**Hint:** Molarity is calculated by dividing the number of moles of solute by the volume of the solution in liters.

### Step 5: Calculate the mass of the solvent
The weight of the solvent (water) can be calculated by subtracting the mass of the solute from the total mass of the solution:
- Mass of solvent = total mass - mass of solute = 100 g - 13 g = 87 g = 0.087 kg

**Hint:** The mass of the solvent can be found by subtracting the mass of the solute from the total mass of the solution.

### Step 6: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
- Molality = moles of solute / mass of solvent in kg = 0.1327 mol / 0.087 kg = 1.52 m

**Hint:** Molality is calculated by dividing the number of moles of solute by the mass of the solvent in kilograms.

### Step 7: Determine the volume to dilute to prepare a 1.5 N solution
To prepare a 1.5 N solution, we need to know the equivalent factor (n-factor) for sulfuric acid, which is 2 (since H₂SO₄ can donate two protons).
- Normality (N) = n-factor × Molarity
- 1.5 N = 2 × Molarity
- Molarity required = 1.5 N / 2 = 0.75 M

### Step 8: Use dilution formula to find the final volume
Using the dilution formula (C1V1 = C2V2):
- C1 = initial molarity = 1.352 M
- V1 = initial volume = 100 mL = 0.1 L
- C2 = final molarity = 0.75 M
- V2 = final volume (unknown)

Rearranging gives:
- V2 = (C1V1) / C2 = (1.352 M × 0.1 L) / 0.75 M = 0.1803 L = 180.3 mL

### Step 9: Calculate the volume of solvent to be added
To find the volume of solvent to be added:
- Volume of solvent to be added = V2 - V1 = 180.3 mL - 100 mL = 80.3 mL

**Hint:** To find the volume of solvent to be added, subtract the initial volume from the final volume.

### Final Answers
- Molarity = 1.352 M
- Molality = 1.52 m
- Volume to which 100 mL of acid should be diluted = 80.3 mL