91Ó°ÊÓ

We are given that the sample has an \(80 \%\) purity. Therefore, the mass of potassium bicarbonate (KHCO3) in the sample can be found by multiplying the initial mass of the sample by the percentage of purity: \(m_{KHCO3} = 1.50 \ \mathrm{g} \times 0.8 = 1.20 \ \mathrm{g}\)

When potassium bicarbonate is heated, it decomposes into potassium carbonate (K2CO3), water (H2O), and carbon dioxide (CO2). The balanced equation for this reaction is: 2 KHCO3 -> K2CO3 + H2O + CO2 Now, we need to find the molar mass of KHCO3, K2CO3, H2O, and CO2. Molar mass of KHCO3 = 39 (K) + 12 (C) + 16x3 (O) + 1 (H) = 100 g/mol Molar mass of K2CO3 = 2x39 (K) + 12 (C) + 3x16 (O) = 138 g/mol Molar mass of H2O = 2x1 (H) + 16 (O) = 18 g/mol Molar mass of CO2 = 12 (C) + 2x16 (O) = 44 g/mol From the balanced equation, we can see that 2 moles of KHCO3 yield 1 mole of K2CO3, 1 mole of H2O, and 1 mole of CO2. So, we can calculate the mass of the remaining residue, K2CO3, after heating, using: \(m_{K2CO3} = (m_{KHCO3} / M_{KHCO3}) \times (1 \mathrm{mole}~K2CO3 / 2 \mathrm{moles}~KHCO3) \times M_{K2CO3}\) \(m_{K2CO3} = (1.20 \ \mathrm{g} / 100 \ \mathrm{g/mol}) \times (1 \mathrm{mole}~K2CO3 / 2 \mathrm{moles}~KHCO3) \times 138 \ \mathrm{g/mol}\) \(m_{K2CO3} \approx 0.828 \ \mathrm{g}\)

We can now find the loss in weight of the sample by subtracting the mass of the remaining residue (K2CO3) from the mass of the initial potassium bicarbonate sample: Loss in weight = \(m_{KHCO3} - m_{K2CO3} = 1.20 \ \mathrm{g} - 0.828 \ \mathrm{g} \approx 0.372 \ \mathrm{g}\) Thus, the loss in weight of the sample, on heating, is approximately \(0.372 \mathrm{~g}\) (option c).