Volume of dil. `HNO_(3)(D=1.1gm//ml,20%3//3)` that can be prepared by diluting 50 ml of concentrated `HNO_(3)(d=1.4gm//ml,70%w//w)` with water is nearest ot :

Answer

To solve the problem of finding the volume of dilute HNO₃ that can be prepared by diluting 50 ml of concentrated HNO₃, we will follow these steps:

### Step-by-Step Solution:

1. **Determine the mass of concentrated HNO₃**:
- Given: Volume of concentrated HNO₃ = 50 ml
- Density of concentrated HNO₃ = 1.4 g/ml
- Mass of concentrated HNO₃ = Volume × Density = 50 ml × 1.4 g/ml = 70 g

2. **Calculate the mass of HNO₃ in the concentrated solution**:
- Concentration of concentrated HNO₃ = 70% w/w
- Mass of HNO₃ = 70% of total mass = 0.70 × 70 g = 49 g

3. **Use the mass of HNO₃ to find the moles of HNO₃**:
- Molar mass of HNO₃ = 63 g/mol
- Moles of HNO₃ = Mass / Molar mass = 49 g / 63 g/mol = 0.7778 mol

4. **Set up the equation for the dilute HNO₃**:
- For dilute HNO₃, we know:
- Density = 1.1 g/ml
- Concentration = 20% w/w
- Let the volume of dilute HNO₃ be \( x \) ml.
- Mass of dilute solution = Density × Volume = 1.1 g/ml × \( x \) ml = 1.1x g
- Mass of HNO₃ in dilute solution = 20% of total mass = 0.20 × 1.1x g = 0.22x g

5. **Equate the moles of HNO₃ from both concentrated and dilute solutions**:
- Moles of HNO₃ in dilute solution = Mass of HNO₃ / Molar mass
- 0.7778 mol = (0.22x g) / (63 g/mol)

6. **Solve for \( x \)**:
- Rearranging gives:
\[ 0.7778 = \frac{0.22x}{63} \]
- Multiply both sides by 63:
\[ 0.7778 \times 63 = 0.22x \]
- Calculate:
\[ 49 = 0.22x \]
- Divide by 0.22:
\[ x = \frac{49}{0.22} \approx 223.64 \text{ ml} \]

7. **Round to the nearest whole number**:
- The volume of dilute HNO₃ that can be prepared is approximately **224 ml**.

### Final Answer:
The volume of dilute HNO₃ that can be prepared is **224 ml**.