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calculate the amount of Calo3 in the following -(1) 16.5 moles of CaCo3 (2) 5×10^46 molecules of CaCo3
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mass of one molecule of caco3 = 40+12+16 = 68u
mass of 1 mole of caco3 = 68 gm
mass of 16.5 mole = 68*16.5 = 1122 gm
6.022*10^23 molecules = 1 mole = 68 gm
1 molecule = 68/6.022*10^23
5*10^46 molecules =( 68/6.022*10^23 )* 5*10^46 = 56.45 * 10^23 gm
mass of 1 mole of caco3 = 68 gm
mass of 16.5 mole = 68*16.5 = 1122 gm
6.022*10^23 molecules = 1 mole = 68 gm
1 molecule = 68/6.022*10^23
5*10^46 molecules =( 68/6.022*10^23 )* 5*10^46 = 56.45 * 10^23 gm
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