The volume of 2N `H_(2)SO_(4)` solution is `0.1dm^(3)`. The volume of its decinormal solution (in `dm^(3)`) will be

Options

A. 0.1
B. 0.2
C. 2
D. 1.7

Answer

To solve the problem, we need to use the relationship between normality and volume for the same substance, which is given by the equation:

\[ N_1 V_1 = N_2 V_2 \]

Where:
- \( N_1 \) = initial normality (2N for \( H_2SO_4 \))
- \( V_1 \) = initial volume (0.1 dm³)
- \( N_2 \) = final normality (decinormal, which is \( \frac{1}{10}N \) or 0.2N)
- \( V_2 \) = final volume (what we need to find)

### Step 1: Identify the known values
- \( N_1 = 2N \)
- \( V_1 = 0.1 \, \text{dm}^3 \)
- \( N_2 = 0.2N \) (since decinormal means \( \frac{1}{10} \) of normal)

### Step 2: Substitute the known values into the equation
Using the equation \( N_1 V_1 = N_2 V_2 \):

\[ 2N \times 0.1 \, \text{dm}^3 = 0.2N \times V_2 \]

### Step 3: Simplify the equation
Now, we can simplify the equation:

\[ 0.2N \, \text{dm}^3 = 0.2N \times V_2 \]

### Step 4: Solve for \( V_2 \)
To find \( V_2 \), we can divide both sides by \( 0.2N \):

\[ V_2 = \frac{0.2N \times 0.1 \, \text{dm}^3}{0.2N} \]

This simplifies to:

\[ V_2 = 2 \, \text{dm}^3 \]

### Final Answer
The volume of the decinormal solution is \( 2 \, \text{dm}^3 \).

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