Answer:
[tex]1.83\times 10^{23}[/tex] number of nitrogen atoms are present
Explanation:
Molar mass of [tex]Ca(NO_{3})_{2}[/tex] is 164.088 g/mol
So, 25.0 g of [tex]Ca(NO_{3})_{2}[/tex] = [tex]\frac{25.0}{164.088}mol[/tex] [tex]Ca(NO_{3})_{2}[/tex] = 0.152 mol of [tex]Ca(NO_{3})_{2}[/tex]
1 mol of atom = [tex]6.023\times 10^{23}[/tex] number of atoms
1 mol of [tex]Ca(NO_{3})_{2}[/tex] contains 2 moles of nitrogen
So, 0.152 moles of [tex]Ca(NO_{3})_{2}[/tex] contain [tex](2\times 0.152)[/tex] moles of nitrogen or 0.304 moles of nitrogen
So, number of nitrogen atoms in 0.304 moles of nitrogen = [tex](0.304\times 6.023\times 10^{23})[/tex] = [tex]1.83\times 10^{23}[/tex]
