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