How did they get 0.49 mass in calculating mole sample

The masses of KHP used in the calculations were:

• Trial 1: $0.21900 \text{ g}$
• Trial 2: $0.28900 \text{ g}$
• Trial 3: $0.22000 \text{ g}$

These masses were then divided by the molar mass of KHP ($204.24 \text{ g/mol}$) to obtain the moles of KHP.

For example, using the mass from Trial 1:

$$\text{Moles of KHP} = \frac{0.21900 \text{ g}}{204.24 \text{ g/mol}} = 0.00107 \text{ mol}$$