20 mL `N/2` HCI, 60 mL `N/10 H_2SO_4` and 150 mL `N/5 HNO_3` are mixed. Calculate the normality of the mixture of acids in solution.

Explanation

To solve the problem of calculating the normality of a mixture of acids, we will follow these steps:

### Step 1: Identify the given data
We have three acids with their respective volumes and normalities:
- Volume of HCl (V1) = 20 mL, Normality (N1) = N/2 = 0.5 N
- Volume of H2SO4 (V2) = 60 mL, Normality (N2) = N/10 = 0.1 N
- Volume of HNO3 (V3) = 150 mL, Normality (N3) = N/5 = 0.2 N

### Step 2: Write the formula for the normality of the mixture
The normality of the mixture (N) can be calculated using the formula:
\[ N = \frac{N_1 \cdot V_1 + N_2 \cdot V_2 + N_3 \cdot V_3}{V_1 + V_2 + V_3} \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ N = \frac{(0.5 \, N) \cdot (20 \, \text{mL}) + (0.1 \, N) \cdot (60 \, \text{mL}) + (0.2 \, N) \cdot (150 \, \text{mL})}{20 \, \text{mL} + 60 \, \text{mL} + 150 \, \text{mL}} \]

### Step 4: Calculate the numerator
Calculating each term in the numerator:
- For HCl: \(0.5 \, N \cdot 20 \, \text{mL} = 10 \, N\)
- For H2SO4: \(0.1 \, N \cdot 60 \, \text{mL} = 6 \, N\)
- For HNO3: \(0.2 \, N \cdot 150 \, \text{mL} = 30 \, N\)

Adding these together:
\[ 10 \, N + 6 \, N + 30 \, N = 46 \, N \]

### Step 5: Calculate the denominator
Calculating the total volume:
\[ V_1 + V_2 + V_3 = 20 \, \text{mL} + 60 \, \text{mL} + 150 \, \text{mL} = 230 \, \text{mL} \]

### Step 6: Calculate the normality of the mixture
Now, substituting back into the formula:
\[ N = \frac{46 \, N}{230 \, \text{mL}} \]

### Step 7: Simplify the expression
Calculating the normality:
\[ N = 0.2 \, N \]

### Final Answer
Thus, the normality of the mixture of acids in solution is **0.2 N**.

---