Answer : The number of atoms present in 159 g of calcium are [tex]23.937\times 10^{23}[/tex]
Explanation :
First we have to calculate the moles of calcium.
[tex]\text{Moles of }Ca=\frac{\text{Mass of }Ca}{\text{Molar mass of }Ca}[/tex]
Molar mass of calcium = 40 g/mole
[tex]\text{Moles of }Ca=\frac{159g}{40g/mole}=3.975mole[/tex]
Now we have to calculate the number of atoms of calcium.
As, 1 mole of calcium contains [tex]6.022\times 10^{23}[/tex] number of atoms of calcium
So, 3.975 mole of calcium contains [tex]3.975\times 6.022\times 10^{23}=23.937\times 10^{23}[/tex] number of atoms of calcium
Therefore, the number of atoms present in 159 g of calcium are [tex]23.937\times 10^{23}[/tex]
