2026 Some Basic Concepts of ChemistryMole Concept and Avogadro Number Medium Question 1 of 2 in Mole Concept and Avogadro Number

The number of hydrogen atoms present in 5.4 g of urea is: (Given: Molar mass of urea: 60 g $mol^{-1}$, $N_A: 6.022 \times 10^{23}$ particles $mol^{-1}$)

A $2.168 \times 10^{23}$
B $2.168 \times 10^{22}$
C $1.084 \times 10^{22}$
D $1.084 \times 10^{23}$

The answer is A.

Solution

Molecular formula of urea = $NH_2CONH_2$; atomicity w.r.t. H = 4 Moles of urea $= \frac{5.4}{60}$ Moles of H $= \frac{5.4}{60} \times 4$ No. of H-atoms $= \frac{5.4}{60} \times 4 \times 6.022 \times 10^{23} = 2.168 \times 10^{23}$

Mole Concept and Avogadro Number in past papers

2 questions from this chapter have appeared across 2 exam years.