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 6N and 2N HCl required to prepare 100 mL of 5N HCl, we can use the concept of normality and the dilution equation. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
We have:
- Normality of solution 1 (N1) = 6N
- Normality of solution 2 (N2) = 2N
- Desired normality of the final solution (N3) = 5N
- Total volume of the final solution (V3) = 100 mL

### Step 2: Set Up the Equation
Using the dilution equation:
\[ N_1 V_1 + N_2 V_2 = N_3 V_3 \]
Where:
- \( V_1 \) = volume of 6N HCl
- \( V_2 \) = volume of 2N HCl

### Step 3: Express the Total Volume
Since the total volume of the mixture is 100 mL, we can express this as:
\[ V_1 + V_2 = 100 \, \text{mL} \]

### Step 4: Substitute Values into the Equation
Substituting the known values into the dilution equation:
\[ 6N \cdot V_1 + 2N \cdot V_2 = 5N \cdot 100 \]

### Step 5: Simplify the Equation
This simplifies to:
\[ 6V_1 + 2V_2 = 500 \]

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

From the first equation, we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = 100 - V_1 \]

Substituting this into the second equation:
\[ 6V_1 + 2(100 - V_1) = 500 \]
\[ 6V_1 + 200 - 2V_1 = 500 \]
\[ 4V_1 + 200 = 500 \]
\[ 4V_1 = 300 \]
\[ V_1 = 75 \, \text{mL} \]

### Step 7: Find \( V_2 \)
Now substituting \( V_1 \) back to find \( V_2 \):
\[ V_2 = 100 - V_1 = 100 - 75 = 25 \, \text{mL} \]

### Final Answer
- Volume of 6N HCl required: **75 mL**
- Volume of 2N HCl required: **25 mL**