0.45 N & 0.6 N NaOH solution are mixed in `2:1` by volume. The amount of solute present in 1 litre of this solution is

Options

A. 0.5 g
B. 25 g
C. 20 g
D. 5 g

Explanation

To find the amount of solute present in 1 litre of the mixed NaOH solution, we can follow these steps:

### Step 1: Determine the volumes of the solutions mixed
We are mixing two solutions of NaOH in a ratio of 2:1 by volume. Let's assume we take:
- Volume of 0.45 N NaOH solution = 2x
- Volume of 0.6 N NaOH solution = x

### Step 2: Calculate the total volume of the mixed solution
The total volume of the mixed solution is:
\[ \text{Total Volume} = 2x + x = 3x \]

### Step 3: Use the normality equation to find the final normality
Using the formula for normality:
\[ N_1V_1 + N_2V_2 = N_fV_f \]
Where:
- \( N_1 = 0.45 \, \text{N} \)
- \( V_1 = 2x \)
- \( N_2 = 0.6 \, \text{N} \)
- \( V_2 = x \)
- \( V_f = 3x \)

Substituting the values:
\[ (0.45)(2x) + (0.6)(x) = N_f(3x) \]
\[ 0.9x + 0.6x = N_f(3x) \]
\[ 1.5x = N_f(3x) \]

### Step 4: Solve for the final normality
Dividing both sides by \( 3x \):
\[ N_f = \frac{1.5x}{3x} = 0.5 \, \text{N} \]

### Step 5: Calculate the amount of solute in 1 litre of the final solution
Normality (N) is defined as the number of equivalents of solute per litre of solution. Therefore, in 1 litre of the final solution:
\[ \text{Number of equivalents} = N_f \times \text{Volume} = 0.5 \times 1 = 0.5 \, \text{equivalents} \]

### Step 6: Calculate the equivalent weight of NaOH
The equivalent weight of NaOH can be calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{N \text{ factor}} = \frac{40 \, \text{g/mol}}{1} = 40 \, \text{g/equiv} \]

### Step 7: Calculate the weight of NaOH in 1 litre of the solution
To find the weight of NaOH in grams:
\[ \text{Weight} = \text{Number of equivalents} \times \text{Equivalent weight} = 0.5 \times 40 = 20 \, \text{grams} \]

### Final Answer
The amount of solute present in 1 litre of this solution is **20 grams** of NaOH.

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