Answer:
\(7.37\times 10^{19}\) number of H atoms are present in 2.00 mg of aspartame
Explanation:
Molecular formula of aspartame is \(C_{14}H_{18}N_{2}O_{5}\)
So, 1 molecule of aspartame contains 18 atoms of H
Molar mass of aspartame = 294.307 g/mol
Number of moles = (mass)/(molar mass)
So, 2.00 mg of aspartame = \(\frac{2.00\times 10^{-3}}{294.307}\) moles of aspartame
We know, 1 mol of a particle = \(6.023\times 10^{23}\) number of particle
So, number of molecules of aspartame in \(\frac{2.00\times 10^{-3}}{294.307}\) moles of aspartame = \(\frac{2.00\times 10^{-3}}{294.307}\times 6.023\times 10^{23}\)
So number of H atoms in \(\frac{2.00\times 10^{-3}}{294.307}\) moles of aspartame = \(\frac{2.00\times 10^{-3}}{294.307}\times 6.023\times 10^{23}\times 18\) = \(7.37\times 10^{19}\)
