Calculation:
To determine the minimum mass of CaO required to remove HCO3− ions completely from 1 kg of hard water, we need to consider the stoichiometry of the reaction between CaO and HCO3− ions.

1. Balanced Chemical Equation:
The reaction between CaO and HCO3− ions can be represented by the following balanced chemical equation:

CaO + HCO3− → CaCO3 + OH−

2. Molar Mass:
The molar mass of CaO is 56.08 g/mol.

3. Conversion Factor:
From the balanced chemical equation, we can see that 1 mole of CaO reacts with 1 mole of HCO3− ions.

4. Molecular Mass of HCO3− ions:
The molecular mass of HCO3− ions can be calculated as follows:
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol (3 oxygen atoms in HCO3− ions)

Molecular mass of HCO3− = 1.01 + 12.01 + (16.00 × 3) = 61.02 g/mol

5. Calculation:
To calculate the minimum mass of CaO required, we need to determine the number of moles of HCO3− ions in 1 kg of the hard water sample.

Given: Concentration of HCO3− ions = 244 ppm
Concentration in g/L = 244 mg/L (since 1 ppm = 1 mg/L)
Concentration in g/mL = 244 μg/mL (since 1 mg = 1000 μg)
Concentration in g/g = 244 μg/g (since 1 mL = 1 g)

Since the density of water is 1 g/mL, the concentration of HCO3− ions in 1 kg of water is equal to 244 μg/g.

Number of moles = Mass / Molar mass = 0.244 mg / 61.02 g/mol = 0.004 mol

From the balanced chemical equation, we know that 1 mole of CaO reacts with 1 mole of HCO3− ions.

Therefore, the minimum mass of CaO required = 0.004 mol × 56.08 g/mol = 0.2243 g

Answer:
The minimum mass of CaO required to remove HCO3− ions completely from 1 kg of the hard water sample is 0.2243 g.