An aqueous solution of urea [6%w/v] is isotonic with NaCI solution then mass-volume percentage (%w/v) of NaCI solution then mass- volume precentage (%w/v) of NaCI solution will be :-

Options

A. `1.46%`
B. `5.85%`
C. `2.92%`
D. `11.7%`

Explanation

To solve the problem of finding the mass-volume percentage (%w/v) of NaCl solution that is isotonic with a 6% w/v urea solution, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Isotonic Solutions**:
- Two solutions are isotonic when they have the same osmotic pressure. This means that the osmotic pressure of the urea solution must equal the osmotic pressure of the NaCl solution.

2. **Use the Formula for Osmotic Pressure**:
- The osmotic pressure (π) can be expressed as:
\[ \pi = i \cdot C \]
- Where \(i\) is the van 't Hoff factor (number of particles the solute dissociates into) and \(C\) is the molarity of the solution.

3. **Identify Values for Urea**:
- For urea, it does not dissociate, so \(i_1 = 1\).
- The concentration \(C_1\) for a 6% w/v solution is calculated as follows:
- 6% w/v means 6 grams of urea in 100 mL of solution.
- Molar mass of urea = 60 g/mol.
- Therefore, the number of moles of urea in 6 grams is:
\[ C_1 = \frac{6 \text{ g}}{60 \text{ g/mol}} = 0.1 \text{ mol} \]
- Since the solution volume is 0.1 L (100 mL), the molarity \(C_1\) is:
\[ C_1 = 0.1 \text{ mol/L} \]

4. **Calculate Osmotic Pressure for Urea**:
- The osmotic pressure for urea is:
\[ \pi_1 = i_1 \cdot C_1 = 1 \cdot 0.1 = 0.1 \text{ osmoles/L} \]

5. **Identify Values for NaCl**:
- For NaCl, it dissociates into two ions, so \(i_2 = 2\).
- Let \(C_2\) be the molarity of the NaCl solution.
- The osmotic pressure for NaCl is:
\[ \pi_2 = i_2 \cdot C_2 = 2 \cdot C_2 \]

6. **Set the Osmotic Pressures Equal**:
- Since the solutions are isotonic:
\[ \pi_1 = \pi_2 \implies 0.1 = 2 \cdot C_2 \]
- Solving for \(C_2\):
\[ C_2 = \frac{0.1}{2} = 0.05 \text{ mol/L} \]

7. **Convert Molarity to %w/v for NaCl**:
- Molar mass of NaCl = 58.5 g/mol.
- To find the mass in grams for 0.05 mol/L in 1000 mL:
\[ \text{Mass of NaCl} = 0.05 \text{ mol/L} \times 58.5 \text{ g/mol} \times 1 \text{ L} = 2.925 \text{ g} \]
- To find the %w/v:
\[ \text{%w/v} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 = \frac{2.925 \text{ g}}{100 \text{ mL}} \times 100 = 2.925\% \]

8. **Final Answer**:
- The mass-volume percentage (%w/v) of the NaCl solution that is isotonic with the 6% w/v urea solution is approximately **2.93%**.