Answer:

To calculate the normality (N) and molarity (M) of the prepared solution, we need to consider the purity of the acid and the dilution factor. The normality is defined as the number of equivalents of solute per liter of solution, while molarity is defined as the number of moles of solute per liter of solution.

Given information:

- Volume of H₃PO₄ taken (V1) = 7.5 mL

- Molecular weight of H₃PO₄ (m.w.) = 98 g/mol

- Density of acid = 1.18 g/cm³

- Purity of the solution = 45% (w/v)

- Volume of the beaker = 2 liters

- Volume of distilled water in the beaker = 1.7 liters

First, let's calculate the mass of H₃PO₄ in the 7.5 mL solution:

\[ \text{Mass of H₃PO₄} = \text{Volume} \times \text{Density} \times \text{Purity} \]

\[ \text{Mass of H₃PO₄} = 7.5 \, \text{mL} \times 1.18 \, \text{g/cm³} \times 0.45 \]

Now, we can calculate the moles of H₃PO₄:

\[ \text{Moles of H₃PO₄} = \frac{\text{Mass of H₃PO₄}}{\text{Molecular weight}} \]

\[ \text{Moles of H₃PO₄} = \frac{\text{Mass of H₃PO₄}}{98} \]

Next, we need to consider the dilution. The total volume after dilution is 2 liters. So, the molarity (M) can be calculated as:

\[ M = \frac{\text{Moles of H₃PO₄}}{\text{Total volume in liters}} \]

Finally, normality (N) can be calculated as normality is twice the molarity for triprotic acids like H₃PO₄:

\[ N = 2 \times M \]

Please plug in the values to get the numerical result.