UCS11B08 Midterm: Question 123
17 Jul 2022
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Q-3) what is the volume (in cm) of a 43. 6 g piece of metal with a density of. The molar mass of al2(co3)3 given is 233. 99 g/mol. According to the question, how many atoms of carbon present in 47. 6 g of al2(co3)3. Step 2: when we break 1 mol of al2(co3)3 we will get 3 mole of carbon atom. Step 3: now let"s calculate how many moles of al2(co3)3 has mass 47. 6 g mole = mass/molar mass. Step 4: previously, we found that 1 mole of al2(co3)3 contains 3 moles of carbon atoms. 0. 203 mole of al2(co3)3 will contain = 3 0. 203. 1 mole of any atom contains 6. 022 1023 atoms. so, 0. 609 mole of carbon atom contains = 0. 609 (6. 022 1023) = 3. 66 1023 number of carbon atom. C = 12 gram/mol, o = 16 gram/mol, h = 1 gram/ mol.
