Question

The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms?

• A. 4 g He
• B. 46 g Na
• C. 0.4 g Ca
• D. 12 g He

Answer 1

The correct option is D 12 g He

Number of moles$=n=\frac{weight}{molecularweight}$

$1.$ $4$ $g$ $He\Rightarrow \frac{4}{4}=1$ $mole$

$2.$ $46$ $g$ $Na\Rightarrow \frac{46}{23}=2$ $moles$

$3.$ $0.4$ $g$ $Ca\Rightarrow \frac{0.4}{40}=0.01$ $moles$

$4.$ $12$ $g$ $He\Rightarrow \frac{12}{4}=3$ $moles$

Number of moles $\propto$ number of atoms

Therefore $12$ $g$ $He$ contains greatest number of atoms.

Hence, option $D$ is correct.