How did they get 0.49 mass in calculating mole sample

Question

How did they get 0.49 mass in calculating mole sample

Asked on March 29, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-03-29T10:50:58.745Z

Based on the provided image and calculations, there is no mass of $0.49 \text{ g}$ used in calculating the moles of potassium hydrogen phthalate (KHP). 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: $$ \text{Moles of KHP} = \frac{0.21900 \text{ g}}{204.24 \text{ g/mol}} = 0.00107 \text{ mol} $$

Accepted Answer · 2026-03-29T10:50:58.745Z

Based on the provided image and calculations, there is no mass of $0.49 \text{ g}$ used in calculating the moles of potassium hydrogen phthalate (KHP). 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: $$ \text{Moles of KHP} = \frac{0.21900 \text{ g}}{204.24 \text{ g/mol}} = 0.00107 \text{ mol} $$