b) 7.50g of a hydrated copper(II) sulphate (CuSO4.nH2O) on heating gave 4.8g of anhydrous copper(II) sulphate, determine the value of n (moles of water of crystallization)

The mass of water lost upon heating is the difference between the mass of the hydrated salt and the anhydrous salt.

$$\text{Mass of } \text{H}_2\text{O} = \text{Mass of hydrated } \text{CuSO}_4 - \text{Mass of anhydrous } \text{CuSO}_4$$ $$\text{Mass of } \text{H}_2\text{O} = 7.50 \text{ g} - 4.80 \text{ g} = 2.70 \text{ g}$$

First, determine the molar mass of $\text{CuSO}_4$.
(Atomic masses: $\text{Cu} = 63.5$, $\text{S} = 32.1$, $\text{O} = 16.0$)

$$\text{Molar mass of } \text{CuSO}_4 = 63.5 + 32.1 + (4 \times 16.0) = 63.5 + 32.1 + 64.0 = 159.6 \text{ g/mol}$$

Now, calculate the moles of $\text{CuSO}_4$.

$$\text{Moles of } \text{CuSO}_4 = \frac{\text{Mass of } \text{CuSO}_4}{\text{Molar mass of } \text{CuSO}_4} = \frac{4.80 \text{ g}}{159.6 \text{ g/mol}} \approx 0.030075 \text{ mol}$$

First, determine the molar mass of $\text{H}_2\text{O}$.
(Atomic masses: $\text{H} = 1.0$, $\text{O} = 16.0$)

$$\text{Molar mass of } \text{H}_2\text{O} = (2 \times 1.0) + 16.0 = 18.0 \text{ g/mol}$$

Now, calculate the moles of $\text{H}_2\text{O}$.

$$\text{Moles of } \text{H}_2\text{O} = \frac{\text{Mass of } \text{H}_2\text{O}}{\text{Molar mass of } \text{H}_2\text{O}} = \frac{2.70 \text{ g}}{18.0 \text{ g/mol}} = 0.15 \text{ mol}$$

The value of $n$ is the ratio of moles of water to moles of anhydrous copper(II) sulfate.

$$n = \frac{\text{Moles of } \text{H}_2\text{O}}{\text{Moles of } \text{CuSO}_4} = \frac{0.15 \text{ mol}}{0.030075 \text{ mol}} \approx 4.987$$

Rounding to the nearest whole number, $n = 5$.

The value of $n$ is $\boxed{\text{5}}$.