`5 mL` of `N-HCl`, `20 mL` of `N//2 H_(2)SO_(4)` and `30 mL` of `N//3` `HNO_(3)` are mixed together and the volume is made to `1 L`. The normality of the resulting solution is

Options

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

Explanation

To find the normality of the resulting solution when mixing different solutions of HCl, H2SO4, and HNO3, we can follow these steps:

### Step 1: Identify the given data
- Volume of HCl (V1) = 5 mL
- Normality of HCl (N1) = N
- Volume of H2SO4 (V2) = 20 mL
- Normality of H2SO4 (N2) = N/2
- Volume of HNO3 (V3) = 30 mL
- Normality of HNO3 (N3) = N/3
- Final volume (Vf) = 1 L = 1000 mL

### Step 2: Use the formula for normality
The formula to find the normality of the resulting solution (Nr) is given by:
\[ N_r \times V_f = N_1 \times V_1 + N_2 \times V_2 + N_3 \times V_3 \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ N_r \times 1000 = N \times 5 + \left(\frac{N}{2}\right) \times 20 + \left(\frac{N}{3}\right) \times 30 \]

### Step 4: Simplify the right-hand side
Calculating each term:
- For HCl: \( N \times 5 = 5N \)
- For H2SO4: \( \left(\frac{N}{2}\right) \times 20 = 10N \)
- For HNO3: \( \left(\frac{N}{3}\right) \times 30 = 10N \)

Now, combine these:
\[ N_r \times 1000 = 5N + 10N + 10N \]
\[ N_r \times 1000 = 25N \]

### Step 5: Solve for the normality of the resulting solution
Now, divide both sides by 1000 to isolate \( N_r \):
\[ N_r = \frac{25N}{1000} \]
\[ N_r = \frac{N}{40} \]

### Final Answer
The normality of the resulting solution is:
\[ N_r = \frac{N}{40} \]

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