Question

The number of atoms of hydrogen present in $12.0{\text{ }}g$of water is :
A.$1.1 \times {10^{23}}$
B.$2.0 \times {10^{23}}$
C.$4.0 \times {10^{23}}$
D.$8.0 \times {10^{23}}$
D.$4.8 \times {10^{23}}$

Answer 1

In this question, calculate the moles of water present. Then simply use the unitary method to find the answer to this question. $6.022 \times {10^{23}}$ is Avogadro's number and represents the number of atoms/molecules present in one mole of the substance.

Complete step by step answer:
One mole water contains one mole of ${\text{O}}$ atoms and two moles of ${\text{H}}$ atoms. We know that one mole or one-gram atom = $6.022 \times {10^{23}}$ atoms. Now let us find the answer to this question:
One mole of ${H_2}O$ ​= one mole of ${\text{O}}$ = $6.022 \times {10^{23}}$ atoms of ${\text{O}}$.
One mole of ${H_2}O$ ​= two moles of ${\text{H}}$=$2 \times 6.022 \times {10^{23}}$atoms = $12.044 \times {10^{23}}$atoms of ${\text{H}}$.
Molar mass of water is $18{\text{ g/mol}}.$
No. of moles in $12.0{\text{ g}}$ =$\dfrac{{12}}{{18}} = 0.67$ moles.
No. of molecules In 1 mole of water = $6.023 \times {10^{23}}\;$ molecules of water.
No. of hydrogen atoms in $0.67$ mole of water= $0.67 \times 12.046 \times {10^{23}} = 6.023 \times {10^{23}} = 8.0 \times {10^{23}}\;$ hydrogen atoms.

Hence, the correct option is option D.

Note:
The concept that a mole of any substance contains the same number of particles was formed out of research which was conducted by Italian physicist Amedeo Avogadro. Avogadro constant can be defined as the number of molecules, atoms, or ions in one mole of a substance: $6.022 \times {10^{23}}$ per mol. It is derived from the number of atoms of the pure isotope $^{{\text{12}}}{\text{C}}$ in 12 grams of that substance and is the reciprocal of atomic mass in grams. The formulae for the mole concept can be written as ${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Mass of the Substance in grams}}}}{{{\text{Molar mass of a Substance}}}} = \dfrac{{{\text{Number of Atoms or Molecules}}}}{{6.022 \times {{10}^{23}}}}$.