Chemistry - Mole Concept Question with Solution | TestHub

Molecular weight of$\mathrm{C}{\mathrm{O}}_2=44,$
Weight = 200 mg = 0.2 g
Number of molecules$=\frac{\mathrm{w}\mathrm{t}.}{\mathrm{m}\mathrm{o}\mathrm{l}.\mathrm{w}\mathrm{t}.}\times {\mathrm{N}}_{\mathrm{A}}$
$=\frac{0.2}{44}\times 6.023\times 10^{23}$
$=2.73\times 10^{21}$
The number of molecules left after the removal of$10^{21}$molecules
$=2.73\times 10^{21}-10^{21}$
$=1.73\times 10^{21}$molecules.