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Q.
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?
Some Basic Concepts of Chemistry
Solution:
(a) $1 $ mol of $He = 4 \,g = N_A$ atoms
or $4\, g$ of $He = \frac{4}{4} mol = 1 N_A$ atom
(b) $1$ mole of $Na = 23\, g = N_A$ atoms
$\quad$ $[\because$ no. of moles $= \frac{\text{given mass}}{\text{atomic mass}}]$
$\therefore 46\,g $ of $ Na= \frac{46}{23}\, mol = \frac{46}{23}N_A$ atoms $= 2N_A$ atoms
(c) $1$ mole of $Ca = 40\, g= N_A$ atoms
$\therefore 0.40\, g $ of $Ca = \frac {0.40}{40}$ mol $ = \frac{0.40}{40} N_A$ atoms
$= 0.01 N_A$ atoms
(d)$1$ mole of $He = 4\, g = N_A$ atoms
$\therefore 12\,g $ of $He = \frac { 12}{4} $ mol $ = \frac { 12}{4} N_A $ atoms $ = 3 N_A$ atoms