The volume of 6 N and 2N HCl required to prepare 100 mL of 5N HCl is

Options

A. `3:1`
B. `1:3`
C. `4:1`
D. `1:4`

Answer

To solve the problem of finding the volumes of 6 N and 2 N HCl required to prepare 100 mL of 5 N HCl, we can follow these steps:

### Step 1: Set Up the Equations
Let:
- \( V_1 \) = volume of 6 N HCl (in mL)
- \( V_2 \) = volume of 2 N HCl (in mL)

From the problem, we know that the total volume of the mixture should be 100 mL:
\[ V_1 + V_2 = 100 \quad \text{(Equation 1)} \]

### Step 2: Use the Normality Equation
The normality equation for mixing solutions is given by:
\[ N_1 V_1 + N_2 V_2 = N_3 V_3 \]
Where:
- \( N_1 = 6 \) N (normality of HCl)
- \( N_2 = 2 \) N (normality of HCl)
- \( N_3 = 5 \) N (desired normality)
- \( V_3 = 100 \) mL (total volume)

Substituting the values into the equation gives:
\[ 6V_1 + 2V_2 = 5 \times 100 \]
\[ 6V_1 + 2V_2 = 500 \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations Simultaneously
Now we have two equations:
1. \( V_1 + V_2 = 100 \)
2. \( 6V_1 + 2V_2 = 500 \)

From Equation 1, we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = 100 - V_1 \]

Substituting \( V_2 \) into Equation 2:
\[ 6V_1 + 2(100 - V_1) = 500 \]
Expanding this gives:
\[ 6V_1 + 200 - 2V_1 = 500 \]
Combining like terms:
\[ 4V_1 + 200 = 500 \]
Subtracting 200 from both sides:
\[ 4V_1 = 300 \]
Dividing by 4:
\[ V_1 = 75 \, \text{mL} \]

### Step 4: Find \( V_2 \)
Using the value of \( V_1 \) to find \( V_2 \):
\[ V_2 = 100 - V_1 = 100 - 75 = 25 \, \text{mL} \]

### Final Answer
Thus, the volumes required to prepare 100 mL of 5 N HCl are:
- Volume of 6 N HCl (\( V_1 \)) = 75 mL
- Volume of 2 N HCl (\( V_2 \)) = 25 mL