Explanation:
We have to find the number of molecules that are present in 0.54 g of Ca(NO₃)₂.
First we have to convert the mass of our sample into moles of Ca(NO₃)₂. We will use the molar mass of Ca(NO₃)₂ to do that.
molar mass of Ca = 40.08 g/mol
molar mass of N = 14.01 g/mol
molar mass of O = 16.00 g/mol
molar mass of Ca(NO₃)₂ = 1 * 40.08 g/mol + 2 * 14.01 g/mol + 6 * 16.00 g/mol
molar mass of Ca(NO₃)₂ = 164.10 g/mol
mass of Ca(NO₃)₂ = 0.54 g
moles of Ca(NO₃)₂ = 0.54 g * 1 mol/(164.10 g)
moles of Ca(NO₃)₂ = 0.00329 moles
According to Avogadro's number there are 6.022 *10^23 molecules in 1 mol of molecules. We can use that relationship to find the number of molecules that are present in our sample.
6.022 *10^23 molecules = 1 mol
molecules of Ca(NO₃)₂ = 0.00329 moles * 6.022 *10^23 molecules/(1 mol)
molecules of Ca(NO₃)₂ = 2.0 * 10^21 molecules
Answer: there are 2.0 * 10^21 molecules of Ca(NO₃)₂ in 0.54 g of it.
