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? - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

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

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 $[∵ no . of moles = \frac{given mass}{atomic mass}]$
$∴ 46 g of Na = \frac{46}{23} mol = \frac{46}{23} N_{A} atoms = 2 N_{A} atoms$
(c) 1 mole of Ca = 40 g = N_A atoms
$∴ 0 .40 g of Ca = \frac{0 .40}{40} mol = \frac{0 .40}{40} N_{A} atoms$
= 0.01 N_A atoms
(d) 1 mole of He = 4 g = N_A atoms
$∴ 12 g of He = \frac{12}{4} mol = \frac{12}{4} N_{A} atoms = 3 N_{A} atoms$

Solution:

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