There are two solutions each at `27^(@)C`
Solution `A`: contains `6g` urea in `200mL` solution
Solution `B`: contains `6g` acetic acid in `100mL` solution.
(i) Are they equimolar ? (ii) Are they isotonic ?

Explanation

To solve the problem, we need to determine whether the two solutions are equimolar and whether they are isotonic.

### Step 1: Calculate the molarity of Solution A (Urea)

1. **Find the number of moles of urea**:
- Given: Mass of urea = 6 g
- Molar mass of urea (NH₂CONH₂) = 60 g/mol
- Number of moles of urea = Mass / Molar mass = 6 g / 60 g/mol = 0.1 mol

2. **Calculate the volume in liters**:
- Volume of solution A = 200 mL = 0.2 L

3. **Calculate molarity (M)**:
- Molarity (M) = Number of moles / Volume in liters = 0.1 mol / 0.2 L = 0.5 M

### Step 2: Calculate the molarity of Solution B (Acetic Acid)

1. **Find the number of moles of acetic acid**:
- Given: Mass of acetic acid = 6 g
- Molar mass of acetic acid (CH₃COOH) = 60 g/mol
- Number of moles of acetic acid = Mass / Molar mass = 6 g / 60 g/mol = 0.1 mol

2. **Calculate the volume in liters**:
- Volume of solution B = 100 mL = 0.1 L

3. **Calculate molarity (M)**:
- Molarity (M) = Number of moles / Volume in liters = 0.1 mol / 0.1 L = 1 M

### Step 3: Compare the molarity of both solutions

- Molarity of Solution A (Urea) = 0.5 M
- Molarity of Solution B (Acetic Acid) = 1 M

**Conclusion for Equimolarity**:
- Since the molarity of Solution A is 0.5 M and the molarity of Solution B is 1 M, they are **not equimolar**.

### Step 4: Determine if the solutions are isotonic

1. **Calculate the osmotic pressure for Solution A (Urea)**:
- For urea (a non-electrolyte), the van 't Hoff factor (i) = 1.
- Osmotic pressure (π) = i * C * R * T
- Here, C = 0.5 M, R = 0.0821 L·atm/(K·mol), T = 300 K (approximately at 27°C).
- π (Urea) = 1 * 0.5 * 0.0821 * 300 = 12.315 atm

2. **Calculate the osmotic pressure for Solution B (Acetic Acid)**:
- For acetic acid, it partially dissociates, but we can approximate i = 2 (considering dimerization).
- Osmotic pressure (π) = i * C * R * T
- Here, C = 1 M.
- π (Acetic Acid) = 2 * 1 * 0.0821 * 300 = 49.26 atm

**Conclusion for Isotonicity**:
- Since the osmotic pressure of Solution A (Urea) is approximately 12.315 atm and the osmotic pressure of Solution B (Acetic Acid) is approximately 49.26 atm, they are **not isotonic**.

### Final Answers:
(i) Are they equimolar? **No**
(ii) Are they isotonic? **No**