Chemistry - Mole Concept Question with Solution | TestHub

$4\mathrm{g}$ of hydrogen $=4$ mole of hydrogen $=4\times 6.023\times 10^{23}$ atoms $$\begin{array}{rl}& 71.0\mathrm{gm}\text{ of chlorine }=\frac{71.0}{71.0}=1\text{ moles of }\\ & \text{ chlorine }\\ & 127\mathrm{gm}\text{ of iodine }=\frac{1}{2}{\text{ mole of }}_2\\ & =\frac{127}{254}=6.023\times 10^{23}\text{ atoms }\\ & 48.0\mathrm{gm}\text{ of magnesium }\\ & =\frac{48.0}{24.0}=2\times 6.023\times 10^{23}\times \frac{1}{2}\text{ atoms }\\ & \therefore 4.0\mathrm{gm}{\mathrm{H}}_2\text{ has largest number of atoms. }\end{array}$$