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 :

Options

A. 150 ml
B. 223 ml
C. 178 ml
D. 333 ml

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 1: Calculate the mass of concentrated HNO₃
Given:
- Volume of concentrated HNO₃ = 50 ml
- Density of concentrated HNO₃ (d) = 1.4 g/ml
- Weight percentage of concentrated HNO₃ = 70% w/w

**Calculation:**
Mass of concentrated HNO₃ solution = Volume × Density
= 50 ml × 1.4 g/ml = 70 g

Now, calculate the mass of HNO₃ in this concentrated solution:
Mass of HNO₃ = 70% of 70 g = 0.70 × 70 g = 49 g

### Step 2: Calculate the moles of concentrated HNO₃
To find the number of moles, we need to use the molar mass of HNO₃, which is approximately 63 g/mol.

**Calculation:**
Moles of HNO₃ = Mass / Molar mass
= 49 g / 63 g/mol ≈ 0.7778 mol

### Step 3: Set up the equation for dilute HNO₃
We want to find the volume of dilute HNO₃ that has a density of 1.1 g/ml and a concentration of 20% w/w.

Let the volume of dilute HNO₃ be \( x \) ml.

**Mass of dilute HNO₃ solution:**
Mass = Volume × Density = \( x \) ml × 1.1 g/ml = 1.1x g

**Mass of HNO₃ in dilute solution:**
Mass of HNO₃ = 20% of total mass
= 0.20 × 1.1x g = 0.22x g

### Step 4: Set the moles of dilute HNO₃ equal to the moles of concentrated HNO₃
Since the moles of HNO₃ remain the same before and after dilution:

0.7778 mol = (0.22x g) / (63 g/mol)

### Step 5: Solve for \( x \)
Rearranging the equation gives:

0.7778 mol × 63 g/mol = 0.22x g

**Calculation:**
49 g = 0.22x g

Now, solving for \( x \):

x = 49 g / 0.22 ≈ 222.73 ml

### Step 6: Round to the nearest whole number
The volume of dilute HNO₃ that can be prepared is approximately 223 ml.

### Final Answer:
The volume of dilute HNO₃ that can be prepared by diluting 50 ml of concentrated HNO₃ is nearest to **223 ml**.

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