Question

How many atoms are in 159g of calcium

Answer 1

(159 g Ca/1) x (1 mol Ca/40.07 g Ca) x (6.02*10^23 atoms Ca/1 mol Ca)
= 2.39*10^24 atoms Ca.

Answer 2

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 $23.937\times 10^{23}$