0.268grams
According to the Avogadro's constant
1 mole of an atom = 6.02 * 10^23 molecules
Given the following
atoms of calcium = 4.02 x 10^21 atoms
Determine the moles of calcium
[tex]\begin{gathered} moles\text{ of Ca}=\frac{4.02\times10^{21}}{6.02\times10^{23}} \\ moles\text{ of Ca}=0.668\times10^{21-23} \\ moles\text{ }of\text{ Ca}=0.668\times10^{-2} \\ moles\text{ of Ca}=0.00688moles \end{gathered}[/tex]Determine the mass of Calcium
[tex]\begin{gathered} Mass=mole\times molar\text{ mass} \\ Mass=0.00688\times40.078 \\ Mass\text{ of Ca}=0.268gram \end{gathered}[/tex]Hence the required grams in 4.02 x 10^21 atoms of calcium is 0.268grams
