To determine the hardness of the water sample prepared by dissolving 1.02 grams of CaCl2 in 750 ml distilled water, we need to calculate the concentration of calcium ions (Ca2+) in the solution, since hardness is usually defined as the concentration of calcium and magnesium ions in water.
The molar mass of CaCl2 is 110.98 g/mol, so 1.02 g of CaCl2 represents:
1.02 g / 110.98 g/mol = 0.00918 mol
When CaCl2 dissolves in water, it dissociates into two ions of Ca2+ and two ions of Cl-. Therefore, the number of moles of Ca2+ ions in the solution is:
0.00918 mol CaCl2 x (1 mol Ca2+ / 1 mol CaCl2) = 0.00918 mol Ca2+
The volume of the solution is 750 ml, which is equivalent to 0.75 L. Therefore, the concentration of Ca2+ ions in the solution is:
0.00918 mol / 0.75 L = 0.0122 M
Finally, we can convert the concentration to units of hardness, which is usually expressed in terms of milligrams of CaCO3 per liter (mg/L). The conversion factor is:
1 mmol CaCO3/L = 100 mg CaCO3/L
Since the molar mass of CaCO3 is 100.09 g/mol, we can calculate the hardness as:
0.0122 mol/L x (1000 mmol/mol) x (100 mg CaCO3/mmol) = 122 mg/L
Therefore, the hardness of the water sample prepared by dissolving 1.02 grams of CaCl2 in 750 ml distilled water is 122 mg/L.