Question

Calculate the number of
(a) Oxygen atoms in 0.10 mole of $N{a}_{2}C{O}_3$. $10H_{2}O$
(b) gram atoms in 4.6 gram of sodium.

Answer 1

(a)
1molecule of $N{a}_{2}C{O}_3.10H_{2}O$ contains oxygen atoms = 13
So, $6.023\times {10}^{23}$ molecules (1mole) has atoms = $13\times 6.023\times {10}^{23}$
So, 0.1 mole will have atoms $=0.1\times 13\times 6.023\times {10}^{23}=7.8\times {10}^{23}$

(b)

$Gramatoms\left(moles\right)=\frac{givenmass}{atomicmass}=\frac{4.6}{23}$ = 0.2 moles or gram atoms