Fill in the missing data in the Table.

Species property	$H_2O$	$CO_2$	$Na$ atom	$MgCl_2$
No. of moles	$2$	-	-	$0.5$
No. of particles	-	$3.011 \times 10^{23}$	-	-
Mass	$36 \, g$	-	$115 \, g$	-

(N/A) To solve this, we use the relations: $\text{Number of particles} = \text{moles} \times 6.022 \times 10^{23}$ and $\text{Mass} = \text{moles} \times \text{Molar mass}$.
$1$. For $CO_2$: Given particles = $3.011 \times 10^{23}$. Moles = $(3.011 \times 10^{23}) / (6.022 \times 10^{23}) = 0.5 \, \text{mol}$. Mass = $0.5 \times 44 \, \text{g/mol} = 22 \, \text{g}$.
$2$. For $Na$ atom: Given mass = $115 \, \text{g}$. Molar mass of $Na = 23 \, \text{g/mol}$. Moles = $115 / 23 = 5 \, \text{mol}$. Particles = $5 \times 6.022 \times 10^{23} = 3.011 \times 10^{24}$.
$3$. For $MgCl_2$: Given moles = $0.5$. Molar mass = $24 + (2 \times 35.5) = 95 \, \text{g/mol}$. Mass = $0.5 \times 95 = 47.5 \, \text{g}$. Particles = $0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$.
$4$. For $H_2O$: Given moles = $2$. Particles = $2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}$.