What is the strength in gram per litre of a solution of `H_(2)SO_(4), 12 mL` of which neutralised by `15 mL` of `N//10` `NaOH` solution?

Explanation

To find the strength in grams per liter of a solution of \( H_2SO_4 \) that neutralizes \( 15 \, \text{mL} \) of \( N/10 \) \( NaOH \), we can follow these steps:

### Step 1: Use the Neutralization Formula
We start with the formula for neutralization:
\[ N_1 V_1 = N_2 V_2 \]
where:
- \( N_1 \) is the normality of \( NaOH \)
- \( V_1 \) is the volume of \( NaOH \) used
- \( N_2 \) is the normality of \( H_2SO_4 \)
- \( V_2 \) is the volume of \( H_2SO_4 \) used

### Step 2: Substitute Known Values
Given:
- \( N_1 = \frac{1}{10} \, \text{N} \) (which is \( 0.1 \, \text{N} \))
- \( V_1 = 15 \, \text{mL} \)
- \( V_2 = 12 \, \text{mL} \)

Substituting these values into the equation:
\[ \left(\frac{1}{10}\right) \times 15 = N_2 \times 12 \]

### Step 3: Solve for \( N_2 \)
Now, we can solve for \( N_2 \):
\[ \frac{15}{10} = N_2 \times 12 \]
\[ 1.5 = N_2 \times 12 \]
\[ N_2 = \frac{1.5}{12} = 0.125 \, \text{N} \]

### Step 4: Calculate Equivalent Mass of \( H_2SO_4 \)
The equivalent mass of \( H_2SO_4 \) is calculated using its molar mass and the n-factor:
- Molar mass of \( H_2SO_4 \) = 98 g/mol
- n-factor for \( H_2SO_4 \) in neutralization = 2

Thus, the equivalent mass is:
\[ \text{Equivalent mass} = \frac{98}{2} = 49 \, \text{g/equiv} \]

### Step 5: Calculate Strength in g/L
Now, we can calculate the strength in grams per liter using the formula:
\[ \text{Strength (g/L)} = N_2 \times \text{Equivalent mass} \times 1000 \]
Substituting the values:
\[ \text{Strength} = 0.125 \times 49 \times 1000 \]
\[ \text{Strength} = 6.125 \, \text{g/L} \]

### Final Answer
The strength of the \( H_2SO_4 \) solution is \( 6.125 \, \text{g/L} \).

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