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

Answer: (D)

$12\, g\, He$

Answer 2

(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