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.
ScanSolve AI Answer
Hey Divine, good to see you again. To calculate the concentration of anhydrous sodium carbonate, we first need its molar mass and the number of moles. Since no specific mass is provided in Table 1, we will assume a mass within the given range of 1.20 g to 1.30 g. Let's assume the mass of sodium carbonate weighed is 1.25 g. Step 1: Calculate the molar mass of anhydrous sodium carbonate (Na_2CO_3). The atomic masses are given as Na = 23, C = 12, O = 16. M(Na_2CO_3) = (2 × M(Na)) + M(C) + (3 × M(O)) M(Na_2CO_3) = (2 × 23) + 12 + (3 × 16) M(Na_2CO_3) = 46 + 12 + 48 M(Na_2CO_3) = 106 g/mol Step 2: Calculate the number of moles of sodium carbonate. We assume the mass of sodium carbonate weighed is 1.25 g. Moles = MassMolar mass Moles of Na_2CO_3 = 1.25 g106 g/mol Moles of Na_2CO_3 ≈ 0.011792 mol Step 3: Convert the volume of the solution to dm^3. The solution is made up to 250 cm^3. 1 dm^3 = 1000 cm^3 Volume = 250 cm^31000 cm^3/dm^3 Volume = 0.250 dm^3 Step 4: Calculate the concentration of sodium carbonate. Concentration = MolesVolume Concentration of Na_2CO_3 = 0.011792 mol0.250 dm^3 Concentration of Na_2CO_3 ≈ 0.047168 mol/dm^3 Rounding to three significant figures: Concentration of Na_2CO_3 ≈ 0.0472 mol/dm^3 The concentration in mol dm^-3 of the anhydrous sodium carbonate is approximately 0.0472 mol/dm^3. Send me the next one 📸