Equal volume of `10% (v//v)` of `HCI` is mixed with `10%(v//v) NaOH` solution. If density of pure `NaOH` is `1.5` times that of pure `HCl` then the resultant solution be ?

Options

A. basic
B. neutral
C. acidic
D. can't be predicted.

Explanation

To solve the problem, we need to determine the nature of the resultant solution when equal volumes of 10% (v/v) HCl and 10% (v/v) NaOH are mixed, given that the density of pure NaOH is 1.5 times that of pure HCl.

### Step-by-Step Solution:

1. **Understanding the Concentration**:
- A 10% (v/v) solution means that there are 10 mL of solute (HCl or NaOH) in every 100 mL of solution.
- Since we are mixing equal volumes, let's assume we take 100 mL of each solution.

2. **Calculating the Volume of Solutes**:
- For HCl: In 100 mL of 10% (v/v) HCl, the volume of HCl is 10 mL.
- For NaOH: In 100 mL of 10% (v/v) NaOH, the volume of NaOH is also 10 mL.

3. **Density Relationship**:
- Let the density of pure HCl be \( d \) g/mL. Then, the density of pure NaOH is \( 1.5d \) g/mL.

4. **Calculating Mass of HCl and NaOH**:
- Mass of HCl in 10 mL = \( 10 \, \text{mL} \times d \, \text{g/mL} = 10d \, \text{g} \).
- Mass of NaOH in 10 mL = \( 10 \, \text{mL} \times 1.5d \, \text{g/mL} = 15d \, \text{g} \).

5. **Calculating Moles of HCl and NaOH**:
- Molar mass of HCl = 1 + 35.5 = 36.5 g/mol.
- Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.
- Moles of HCl = \( \frac{10d}{36.5} \).
- Moles of NaOH = \( \frac{15d}{40} \).

6. **Calculating Moles**:
- Moles of HCl = \( \frac{10d}{36.5} \approx 0.274d \).
- Moles of NaOH = \( \frac{15d}{40} = 0.375d \).

7. **Comparing Moles**:
- Since \( 0.375d > 0.274d \), it shows that there are more moles of NaOH than HCl.

8. **Determining the Nature of the Resultant Solution**:
- Since NaOH is a strong base and there are more moles of NaOH than HCl, the resultant solution will be basic.

### Conclusion:
The resultant solution is basic.