Sure, let's solve this problem step by step.
1. Determine the mass of water lost during heating:
When the hydrated copper sulphate is heated, it loses water. The mass of water lost can be calculated by subtracting the mass of the anhydrous copper sulphate from the mass of the hydrated copper sulphate.
[tex]\[
\text{Mass of water lost} = \text{Mass of hydrated copper sulphate} - \text{Mass of anhydrous copper sulphate}
\][/tex]
Given:
[tex]\[
\text{Mass of hydrated copper sulphate} = 11.25 \, \text{g}
\][/tex]
[tex]\[
\text{Mass of anhydrous copper sulphate} = 7.19 \, \text{g}
\][/tex]
Therefore:
[tex]\[
\text{Mass of water lost} = 11.25 \, \text{g} - 7.19 \, \text{g} = 4.06 \, \text{g}
\][/tex]
2. Calculate the moles of anhydrous copper sulphate [tex]\( \text{CuSO}_4 \)[/tex]:
To find the moles of anhydrous copper sulphate, we use its molar mass. The molar mass of [tex]\(\text{CuSO}_4\)[/tex] is:
[tex]\[
\text{Molar mass of CuSO}_4 = 159.61 \, \text{g/mol}
\][/tex]
Using the formula [tex]\( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \)[/tex], we get:
[tex]\[
\text{Moles of CuSO}_4 = \frac{7.19 \, \text{g}}{159.61 \, \text{g/mol}} = 0.0450473028005764 \, \text{mol}
\][/tex]
3. Calculate the moles of water lost [tex]\( \text{H}_2\text{O} \)[/tex]:
To find the moles of water, we use the molar mass of water. The molar mass of [tex]\(\text{H}_2\text{O}\)[/tex] is:
[tex]\[
\text{Molar mass of H}_2\text{O} = 18.015 \, \text{g/mol}
\][/tex]
Thus:
[tex]\[
\text{Moles of H}_2\text{O} = \frac{4.06 \, \text{g}}{18.015 \, \text{g/mol}} = 0.22536774909797389 \, \text{mol}
\][/tex]
4. Determine the value of [tex]\( x \)[/tex]:
The value of [tex]\( x \)[/tex] is the ratio of the moles of water to the moles of anhydrous copper sulphate.
[tex]\[
x = \frac{\text{Moles of H}_2\text{O}}{\text{Moles of CuSO}_4} = \frac{0.22536774909797389 \, \text{mol}}{0.0450473028005764 \, \text{mol}} = 5.002913273091463
\][/tex]
Rounding [tex]\( x \)[/tex] to the nearest whole number, we find:
[tex]\[
x \approx 5
\][/tex]
Therefore, the value of [tex]\( x \)[/tex] in the hydrated copper sulphate [tex]\(\text{CuSO}_4 \cdot xH_2O\)[/tex] is 5.
