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* 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=(159g)/(40g/mole)=3.975mole`

Now we have to calculate the number of atoms of calcium.

As, 1 mole of calcium contains
`6.022* 10^(23)` number of atoms of calcium

So, 3.975 mole of calcium contains
`3.975* 6.022* 10^(23)=23.937* 10^(23)` number of atoms of calcium

Therefore, the number of atoms present in 159 g of calcium are
`23.937* 10^(23)`