
The molecular formula of ammonia is $N{H_3}$ . Therefore, the number of hydrogen atoms in $34g$ of ammonia will be:
A. $3 \times 6.023 \times {10^{23}}$
B. $6.023 \times {10^{23}}$
C. $34 \times 6.023 \times {10^{23}}$
D. $6 \times 6.023 \times {10^{23}}$
Answer
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Hint: The number of particles, atoms, ions, molecules, etc present in one mole of any substance is always equal to the Avogadro number whose value is equal to $6.023 \times {10^{23}}$ . It is important to determine how many atoms of an element is present in a compound.
Complete step by step answer: Ammonia is a compound consisting of 1 atom of nitrogen attached with 3 atoms of hydrogen.
The atomic weight of nitrogen and hydrogen is defined as the weight of one atom of the corresponding element.
Atomic weight of nitrogen = $14g$
Atomic weight of hydrogen = $1g$
Molecular weight of ammonia whose chemical formula is $N{H_3}$ , will be:
$N{H_3} = (1 \times 14) + (3 \times 1) = 17g$
We also know that one mole of any compound is equal to the molecular weight of the compound. So, $17g$ of ammonia will comprise $1mol$ .
As per the question, we have $34g$ of ammonia.
Thus, according to the mathematical formula of number of moles, we have:
$n = \dfrac{w}{{{M_w}}}$
Where, $n = $ number of moles
$w = $ given weight of compound
${M_w} = $ Molecular weight of compound
Thus, $n = \dfrac{{34}}{{17}} = 2 moles$
As we can see, one mole of ammonia has 3 moles of hydrogen atoms in it. So, two mole of ammonia will have:
${n_H} = 2 \times 3 = 6mol$
We also know that the number of particles in one mole of a substance is equal to the Avogadro number.
Hence, the number of hydrogen atoms in six moles of hydrogen will be:
${N_H} = 6 \times 6.023 \times {10^{23}}$
Thus, the correct option is D. $6 \times 6.023 \times {10^{23}}$.
Note: Ammonia is a colorless gas with a pungent smell. It is produced naturally in the human body and in the nature with the help of many nitrogen fixing bacteria in water, air and mainly in soil to form ammonium ions which get oxidized to form nitrates and nitrites which are the products that are used by the plants for their food synthesis. The stoichiometry of ammonia is very crucial in many industrial processes and is mainly generated in industries with the help of Haber’s process at high temperature and high pressure.
Complete step by step answer: Ammonia is a compound consisting of 1 atom of nitrogen attached with 3 atoms of hydrogen.
The atomic weight of nitrogen and hydrogen is defined as the weight of one atom of the corresponding element.
Atomic weight of nitrogen = $14g$
Atomic weight of hydrogen = $1g$
Molecular weight of ammonia whose chemical formula is $N{H_3}$ , will be:
$N{H_3} = (1 \times 14) + (3 \times 1) = 17g$
We also know that one mole of any compound is equal to the molecular weight of the compound. So, $17g$ of ammonia will comprise $1mol$ .
As per the question, we have $34g$ of ammonia.
Thus, according to the mathematical formula of number of moles, we have:
$n = \dfrac{w}{{{M_w}}}$
Where, $n = $ number of moles
$w = $ given weight of compound
${M_w} = $ Molecular weight of compound
Thus, $n = \dfrac{{34}}{{17}} = 2 moles$
As we can see, one mole of ammonia has 3 moles of hydrogen atoms in it. So, two mole of ammonia will have:
${n_H} = 2 \times 3 = 6mol$
We also know that the number of particles in one mole of a substance is equal to the Avogadro number.
Hence, the number of hydrogen atoms in six moles of hydrogen will be:
${N_H} = 6 \times 6.023 \times {10^{23}}$
Thus, the correct option is D. $6 \times 6.023 \times {10^{23}}$.
Note: Ammonia is a colorless gas with a pungent smell. It is produced naturally in the human body and in the nature with the help of many nitrogen fixing bacteria in water, air and mainly in soil to form ammonium ions which get oxidized to form nitrates and nitrites which are the products that are used by the plants for their food synthesis. The stoichiometry of ammonia is very crucial in many industrial processes and is mainly generated in industries with the help of Haber’s process at high temperature and high pressure.
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