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Which sample has the largest number of atoms?
(a)$1mg{\text{ of }}{C_4}{H_{10}}$
(b) $1mg{\text{ of }}{N_2}$
(c) $1mg{\text{ of }}Na$
(d) $1mL{\text{ of }}{H_2}O$

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Last updated date: 20th May 2024
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Answer
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Hint: In periodic table the atomic weight of each element has been listed and this helps scientists to describe the mass of atoms. The sample of an element is weighed in grams and we can calculate the number of atoms if we have information of three pieces- atomic weight, grams and Avogadro’s number.

Complete answer:
First, calculate the total number of atoms present in the given sample and multiply it with Avogadro’s number $(N)$ and divide it by the molecular mass of the sample.
$1mg{\text{ of }}{C_4}{H_{10}} = 14 \times N \times {10^{ - 3}}{(58)^{ - 1}} = 1.45 \times {10^{20}}$ atoms.
$1mg{\text{ of }}{N_2} = 2N \times {10^{ - 3}}{(28)^{ - 1}} = 0.43 \times {10^{20}}$ atoms.
$1mg{\text{ of }}Na = N \times {10^{ - 3}}{(23)^{ - 1}} = 0.26 \times {10^{20}}$ atoms
$1mL = 1g{\text{ of }}{H_2}O = 3N{(18)^{ - 1}} = 1 \times {10^{23}}$ atoms. (Here $N$ is Avogadro’s number)
Therefore, option (D) $1mL{\text{ of }}{H_2}O$ has the largest number of atoms.

Note:
To find the largest number of atoms we need to remember the molecular mass of the sample. The total number of atoms present in a molecule is calculated by adding up the atoms present in them and the Avogadro’s number $(N)$ is $6.02 \times {10^{23}}$.
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