Question

Calculate the number of moles, molecules and total number of atoms present in $1.1\times {10}^{-4}kg$ of $C{O}_2$.
[Given: $C=12,O=16,N=14,H=1,He=4$]

Answer 1

Mass of $C{O}_2=1.1\times {10}^{-4}Kg$

Molar mass of $C{O}_2=12+32$

$\Rightarrow 44gmo{l}^{-1}$

$\Rightarrow 44\times {10}^{-3}Kgmo{l}^{-1}$

(i) Number of moles (n)= $\frac{MassofC{O}_2}{MolarmassofC{O}_2}$

$=\frac{1.1\times {10}^{-4}Kg}{44\times {10}^{-3}Kgmo{l}^{-1}}$

$\Rightarrow n=2.5\times {10}^{-3}mol$

(ii) Number of molecules of $C{O}_2=n\times$ Avogadro's number

$=2.5\times {10}^{-3}\times 6.022\times {10}^{23}$

$=1.5\times {10}^{21}$ molecules

(iii) Number of atoms present= $1.5\times {10}^{21}\times 3$

$=4.5\times {10}^{21}$ atoms

$C{O}_2$ is triatomic, hence multiplied by $3$.