`5 ml` of `HCl ,20 ml` of `N//2` of `H_(2)SO_(4)` and `30 ml` of `N//3 HNO_(3)` are mixed together and volume made to one litre.

Options

A. ` (N)/(40) `
B. ` (N)/(10) `
C. ` (N)/(20) `
D. ` (N)/(5) `

Explanation

To find the normality of the new solution formed by mixing 5 ml of HCl, 20 ml of N/2 H₂SO₄, and 30 ml of N/3 HNO₃ and making the total volume 1 liter, we can follow these steps:

### Step 1: Identify the normalities and volumes of the solutions
- **HCl**: Normality (N₁) = 1 N, Volume (V₁) = 5 ml
- **H₂SO₄**: Normality (N₂) = N/2 = 0.5 N, Volume (V₂) = 20 ml
- **HNO₃**: Normality (N₃) = N/3 ≈ 0.33 N, Volume (V₃) = 30 ml

### Step 2: Convert volumes to liters
- V₁ = 5 ml = 0.005 L
- V₂ = 20 ml = 0.020 L
- V₃ = 30 ml = 0.030 L

### Step 3: Use the formula for normality
The formula for calculating the normality of the mixed solution is given by:
\[ N_{final} \times V_{final} = N_1 \times V_1 + N_2 \times V_2 + N_3 \times V_3 \]
Where \( V_{final} \) is the final volume of the solution (1 L = 1000 ml).

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ N_{final} \times 1 = (1 \, \text{N} \times 0.005 \, \text{L}) + (0.5 \, \text{N} \times 0.020 \, \text{L}) + (0.33 \, \text{N} \times 0.030 \, \text{L}) \]

### Step 5: Calculate each term
1. For HCl:
\[ 1 \times 0.005 = 0.005 \]
2. For H₂SO₄:
\[ 0.5 \times 0.020 = 0.010 \]
3. For HNO₃:
\[ 0.33 \times 0.030 \approx 0.0099 \]

### Step 6: Sum the contributions
Now, sum the contributions:
\[ 0.005 + 0.010 + 0.0099 = 0.0249 \]

### Step 7: Solve for \( N_{final} \)
Now, substituting back into the equation:
\[ N_{final} \times 1 = 0.0249 \]
Thus,
\[ N_{final} = 0.0249 \, \text{N} \]

### Step 8: Convert to a more usable form
To express this in terms of a fraction, we can multiply by 40 to find a more standard normality:
\[ N_{final} \approx \frac{0.0249}{1} \times 40 \approx 1 \]

### Final Answer
The normality of the new solution is approximately **0.025 N** or **1/40 N**.