Maximum number of atoms are present in which of the following?
(1) 2g atom Fe
(2) 2g molecule ${H_2}O$
(3) 16g $C{H_4}$
(4) 10L ${H_2}O$ (l)
Answer
597.9k+ views
Hint:Calculate the number of moles present in each of the given compounds using their molecular weights and given weights. One mole of a substance consists of Avogadro number of molecules and hence the number of atoms can be calculated accordingly.
Complete answer:
(1) Given 2 grams of Iron atoms. The atomic weight of Iron is $56$ grams. So, $56$ grams of Iron contains Avogadro number i.e, $6.022 \times {10^{23}}$ atoms.
Hence 2 grams of Iron contains $\dfrac{{6.022 \times {{10}^{23}}}}{{56}} \times 2$
This gives $0.215 \times {10^{23}}$ atoms.
(2) Molecular weight of ${H_2}O$ is $18$ grams. Two grams of ${H_2}O$ is given to us.
We can calculate the number of moles as $\dfrac{2}{{18}} = \dfrac{1}{9}$ . One mole consists of $6.022 \times {10^{23}}$ number of molecules and each molecule contains three atoms.
Therefore, the total number of atoms would be $6.022 \times {10^{23}} \times \dfrac{1}{9} \times 3 = 2.006 \times {10^{23}}$
(3) Molecular weight of $C{H_4}$ is $16$ grams. Given to us are $16$ grams. Hence there is one mole of $C{H_4}$ and one mole contains $6.022 \times {10^{23}}$ molecules.
We can calculate the total number of atoms to be $5 \times 6.022 \times {10^{23}} = 30.11 \times {10^{23}}$
(4) We know that $22.4$ liters is one mole. We can calculate the number of moles for $10$ liters to be $\dfrac{{10 \times 1mole}}{{22.4}} = 0.446$ moles.
Number of molecules would be $0.446 \times 6.022 \times {10^{23}}$ which is approximately $2.68 \times {10^{23}}$ molecules.
One molecule consists of three atoms so the number of atoms will be $3 \times 2.68 \times {10^{23}} = 8.04 \times {10^{23}}$
Clearly, we can say that $16g$ of $C{H_4}$ contains the maximum number of atoms.
So the correct answer is Option 3.
Note:
One mole of an atom or a molecule contains Avogadro number of atoms or molecules. If it is a molecule, then one should not forget to change it to atoms by multiplying it by the number of atoms present in one individual molecule.
Complete answer:
(1) Given 2 grams of Iron atoms. The atomic weight of Iron is $56$ grams. So, $56$ grams of Iron contains Avogadro number i.e, $6.022 \times {10^{23}}$ atoms.
Hence 2 grams of Iron contains $\dfrac{{6.022 \times {{10}^{23}}}}{{56}} \times 2$
This gives $0.215 \times {10^{23}}$ atoms.
(2) Molecular weight of ${H_2}O$ is $18$ grams. Two grams of ${H_2}O$ is given to us.
We can calculate the number of moles as $\dfrac{2}{{18}} = \dfrac{1}{9}$ . One mole consists of $6.022 \times {10^{23}}$ number of molecules and each molecule contains three atoms.
Therefore, the total number of atoms would be $6.022 \times {10^{23}} \times \dfrac{1}{9} \times 3 = 2.006 \times {10^{23}}$
(3) Molecular weight of $C{H_4}$ is $16$ grams. Given to us are $16$ grams. Hence there is one mole of $C{H_4}$ and one mole contains $6.022 \times {10^{23}}$ molecules.
We can calculate the total number of atoms to be $5 \times 6.022 \times {10^{23}} = 30.11 \times {10^{23}}$
(4) We know that $22.4$ liters is one mole. We can calculate the number of moles for $10$ liters to be $\dfrac{{10 \times 1mole}}{{22.4}} = 0.446$ moles.
Number of molecules would be $0.446 \times 6.022 \times {10^{23}}$ which is approximately $2.68 \times {10^{23}}$ molecules.
One molecule consists of three atoms so the number of atoms will be $3 \times 2.68 \times {10^{23}} = 8.04 \times {10^{23}}$
Clearly, we can say that $16g$ of $C{H_4}$ contains the maximum number of atoms.
So the correct answer is Option 3.
Note:
One mole of an atom or a molecule contains Avogadro number of atoms or molecules. If it is a molecule, then one should not forget to change it to atoms by multiplying it by the number of atoms present in one individual molecule.
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