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. $4gHe$
• B. $46gNa$
• C. $0.4gCa$
• D. $12gHe$

Answer 1

Answer: (D)

(a) $1$ mol of $He=4g=N_A$ atoms
or $4g$ of $He=\frac{4}{4}mol=1N_A$ atom
(b) $1$ mole of $Na=23g=N_A$ atoms
$[\because$ no. of moles $=\frac{\text{given mass}}{\text{atomic mass}}]$
$\therefore 46g$ of $Na=\frac{46}{23}mol=\frac{46}{23}{N}_A$ atoms $=2N_A$ atoms
(c) $1$ mole of $Ca=40g=N_A$ atoms
$\therefore 0.40g$ of $Ca=\frac{0.40}{40}$ mol $=\frac{0.40}{40}{N}_A$ atoms
$=0.01N_A$ atoms
(d)$1$ mole of $He=4g=N_A$ atoms
$\therefore 12g$ of $He=\frac{12}{4}$ mol $=\frac{12}{4}{N}_A$ atoms $=3N_A$ atoms