What will be the normality of a solution containing 4.9 g `H_(3) PO_(4)` dissolved in 500 ml water ?

Options

A. `0.3`
B. `1.0`
C. `3.0`
D. `0.1`

Explanation

To find the normality of a solution containing 4.9 g of \( H_3PO_4 \) dissolved in 500 mL of water, we will follow these steps:

### Step 1: Calculate the number of moles of \( H_3PO_4 \)

To calculate the number of moles, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of \( H_3PO_4 \) can be calculated as follows:

- Hydrogen (H): 1 g/mol × 3 = 3 g/mol
- Phosphorus (P): 31 g/mol × 1 = 31 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Adding these together gives:

\[ \text{Molar mass of } H_3PO_4 = 3 + 31 + 64 = 98 \text{ g/mol} \]

Now, substituting the values into the moles formula:

\[ \text{Number of moles} = \frac{4.9 \text{ g}}{98 \text{ g/mol}} = 0.0500 \text{ moles} \]

### Step 2: Convert the volume of the solution from mL to L

The volume of the solution is given as 500 mL. To convert this to liters:

\[ \text{Volume in L} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L} \]

### Step 3: Calculate the molarity of the solution

Molarity (M) is calculated using the formula:

\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume of solution in L}} \]

Substituting the values:

\[ \text{Molarity} = \frac{0.0500 \text{ moles}}{0.5 \text{ L}} = 0.100 \text{ M} \]

### Step 4: Determine the n-factor of \( H_3PO_4 \)

The n-factor for \( H_3PO_4 \) (phosphoric acid) is equal to its basicity, which is the number of replaceable hydrogen ions. \( H_3PO_4 \) can donate 3 hydrogen ions, so:

\[ \text{n-factor} = 3 \]

### Step 5: Calculate the normality of the solution

Normality (N) is calculated using the formula:

\[ \text{Normality} = \text{Molarity} \times \text{n-factor} \]

Substituting the values:

\[ \text{Normality} = 0.100 \text{ M} \times 3 = 0.300 \text{ N} \]

### Final Answer

The normality of the solution is:

\[ \text{Normality} = 0.300 \text{ N} \text{ (or 0.3 N)} \]