Consider the following aqueous solutions
I. 6 g urea in 100 ml
II. 6 g acetic acid in 100 mL solution
III . 6g propanol in 100 mL solution
Which has minimum molarity of solute at equilibrium ?

Options

A. Only I
B. Only II
C. Only III
D. Equal

Answer

To determine which solution has the minimum molarity of solute at equilibrium, we will calculate the molarity for each of the given solutions: urea, acetic acid, and propanol.

### Step 1: Calculate the Molarity of Urea
1. **Identify the mass of urea**: 6 g
2. **Calculate the molar mass of urea (NH₂CONH₂)**:
- Nitrogen (N): 14 g/mol (2 atoms) = 2 × 14 = 28 g/mol
- Hydrogen (H): 1 g/mol (4 atoms) = 4 × 1 = 4 g/mol
- Carbon (C): 12 g/mol (1 atom) = 12 g/mol
- Oxygen (O): 16 g/mol (1 atom) = 16 g/mol
- Total molar mass = 28 + 4 + 12 + 16 = 60 g/mol
3. **Calculate the number of moles of urea**:
\[ \text{Moles of urea} = \frac{\text{mass}}{\text{molar mass}} = \frac{6 \text{ g}}{60 \text{ g/mol}} = 0.1 \text{ mol} \]
4. **Convert volume from mL to L**: 100 mL = 0.1 L
5. **Calculate molarity**:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.1 \text{ mol}}{0.1 \text{ L}} = 1 \text{ M} \]

### Step 2: Calculate the Molarity of Acetic Acid
1. **Identify the mass of acetic acid**: 6 g
2. **Calculate the molar mass of acetic acid (CH₃COOH)**:
- Carbon (C): 12 g/mol (2 atoms) = 2 × 12 = 24 g/mol
- Hydrogen (H): 1 g/mol (4 atoms) = 4 × 1 = 4 g/mol
- Oxygen (O): 16 g/mol (2 atoms) = 2 × 16 = 32 g/mol
- Total molar mass = 24 + 4 + 32 = 60 g/mol
3. **Calculate the number of moles of acetic acid**:
\[ \text{Moles of acetic acid} = \frac{6 \text{ g}}{60 \text{ g/mol}} = 0.1 \text{ mol} \]
4. **Convert volume from mL to L**: 100 mL = 0.1 L
5. **Calculate molarity**:
\[ \text{Molarity (M)} = \frac{0.1 \text{ mol}}{0.1 \text{ L}} = 1 \text{ M} \]

### Step 3: Calculate the Molarity of Propanol
1. **Identify the mass of propanol**: 6 g
2. **Calculate the molar mass of propanol (C₃H₈O)**:
- Carbon (C): 12 g/mol (3 atoms) = 3 × 12 = 36 g/mol
- Hydrogen (H): 1 g/mol (8 atoms) = 8 × 1 = 8 g/mol
- Oxygen (O): 16 g/mol (1 atom) = 16 g/mol
- Total molar mass = 36 + 8 + 16 = 60 g/mol
3. **Calculate the number of moles of propanol**:
\[ \text{Moles of propanol} = \frac{6 \text{ g}}{60 \text{ g/mol}} = 0.1 \text{ mol} \]
4. **Convert volume from mL to L**: 100 mL = 0.1 L
5. **Calculate molarity**:
\[ \text{Molarity (M)} = \frac{0.1 \text{ mol}}{0.1 \text{ L}} = 1 \text{ M} \]

### Conclusion
The molarity of all three solutions (urea, acetic acid, and propanol) is 1 M. Therefore, none of the solutions has a minimum molarity at equilibrium; they all have the same molarity.