6% (W/V) solution of urea will be isotonic with:

Options

A. 18% (W/V) solution of glucose
B. 0.5 M solution of NaCl
C. 1 M solution of `CH_(3) CO OH`
D. 6% (W/V) solution of sucrose.

Answer

To determine which solution is isotonic with a 6% (W/V) solution of urea, we need to calculate the molarity of the urea solution and compare it with the molarity of the other solutions provided.

### Step-by-Step Solution:

1. **Understanding the 6% (W/V) Solution**:
- A 6% (W/V) solution means there are 6 grams of urea in 100 mL of solution.

2. **Calculating Molarity of Urea**:
- Molarity (M) is defined as the number of moles of solute per liter of solution.
- First, we need to calculate the number of moles of urea:
\[ \text{Number of moles} = \frac{\text{Weight (g)}}{\text{Molecular weight (g/mol)}} \]
- The molecular weight of urea (NH₂CONH₂) is 60 g/mol.
- Therefore, the number of moles of urea in 6 grams is:
\[ \text{Number of moles} = \frac{6 \text{ g}}{60 \text{ g/mol}} = 0.1 \text{ moles} \]

3. **Converting Volume to Liters**:
- The volume of the solution is 100 mL, which is equivalent to 0.1 L.

4. **Calculating Molarity**:
- Now, we can calculate the molarity of the urea solution:
\[ M = \frac{\text{Number of moles}}{\text{Volume (L)}} = \frac{0.1 \text{ moles}}{0.1 \text{ L}} = 1 \text{ M} \]

5. **Identifying Isotonic Solutions**:
- A solution is isotonic if it has the same osmotic pressure, which can be achieved if the molarity of solute particles is the same.
- Urea does not dissociate in solution, so its van 't Hoff factor (i) is 1.

6. **Calculating Molarity of Other Solutions**:
- **Glucose (18% W/V)**:
- Molecular weight of glucose (C₆H₁₂O₆) = 180 g/mol.
- Molarity calculation:
\[ \text{Molarity} = \frac{18 \text{ g}}{180 \text{ g/mol}} \times \frac{1000}{100} = 1 \text{ M} \]
- **NaCl (0.5 M)**:
- NaCl dissociates into Na⁺ and Cl⁻, so:
\[ \text{Effective molarity} = 0.5 \text{ M} \times 2 = 1 \text{ M} \]
- **Acetic Acid (1 M)**:
- Acetic acid is an organic compound and does not dissociate significantly, so it remains 1 M.
- **Sucrose (6% W/V)**:
- Molecular weight of sucrose (C₁₂H₂₂O₁₁) = 342 g/mol.
- Molarity calculation:
\[ \text{Molarity} = \frac{6 \text{ g}}{342 \text{ g/mol}} \times \frac{1000}{100} \approx 0.175 \text{ M} \]
- This is not isotonic as it is less than 1 M.

### Conclusion:
The solutions that are isotonic with the 6% (W/V) urea solution are:
- Glucose (18% W/V)
- NaCl (0.5 M)
- Acetic Acid (1 M)