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. $0.4gCa$
• D. 12 g He

Answer 1

Answer: (D)

12 g He

Answer 2

(4) : (1) 1 mol of He = 4 g = $N_{A}$atom

or 4 g of He $= \frac{4}{4}mol = 1N_{A}$atom

(2) 1 mole of Na = 23 g = $1N_{A}$atom

$\therefore$no. of moles

=$\frac{givenmass}{Atomicmass}$

$\therefore$46 g of Na $= \frac{46}{23}mol = \frac{46}{23}N_{A}$atoms

$= 2N_{A}$atoms

(3) 1 mole of Ca = 40 g =$N_{A}$atoms

$\therefore 0.40gofCa = \frac{0.40}{40}mol = \frac{0.40}{40}N_{A}$atoms

$= 0.01N_{A}$atoms

(4) 1 mole of He = 4g=$N_{A}$atoms

$\therefore$12 g of He = $\frac{12}{4}mol = \frac{12}{4}N_{A}$atoms

$= 3N_{A}$atoms