A dilute solution of `H_(2)SO_(4)` is made by adding 5 mL of `3N H_(2)SO_(4)` to 245 mL of water. Find the normality and molarity of the diluted solution.

Answer

To solve the problem of finding the normality and molarity of the diluted solution of `H2SO4`, we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- Volume of `H2SO4` solution (V1) = 5 mL
- Normality of `H2SO4` solution (N1) = 3 N
- Volume of water added (V2) = 245 mL

### Step 2: Calculate the Total Volume of the Diluted Solution
The total volume of the diluted solution (V_total) is the sum of the volume of `H2SO4` added and the volume of water:
\[ V_{total} = V1 + V2 = 5 \, \text{mL} + 245 \, \text{mL} = 250 \, \text{mL} \]

### Step 3: Convert the Total Volume to Liters
Since normality is typically expressed in terms of liters, we need to convert the total volume from mL to L:
\[ V_{total} = 250 \, \text{mL} = 0.250 \, \text{L} \]

### Step 4: Calculate the Normality of the Diluted Solution (N2)
Using the dilution formula for normality:
\[ N2 = \frac{N1 \times V1}{V_{total}} \]
Substituting the values:
\[ N2 = \frac{3 \, \text{N} \times 5 \, \text{mL}}{250 \, \text{mL}} \]
\[ N2 = \frac{15}{250} \]
\[ N2 = 0.0612 \, \text{N} \]

### Step 5: Calculate the Basicity of `H2SO4`
The basicity of `H2SO4` (the number of replaceable hydrogen ions) is 2, as it can donate two protons (H⁺).

### Step 6: Calculate the Molarity of the Diluted Solution (M)
Using the relationship between normality, molarity, and basicity:
\[ M = \frac{N}{\text{Basicity}} \]
Substituting the known values:
\[ M = \frac{0.0612 \, \text{N}}{2} \]
\[ M = 0.0306 \, \text{M} \]

### Final Results
- Normality (N2) of the diluted solution = 0.0612 N
- Molarity (M) of the diluted solution = 0.0306 M

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