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# Calculate the number of oxygen atoms in $0.2$ mole of $N{a_2}C{O_3} \cdot 10{H_2}O$ .

Last updated date: 25th Jul 2024
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Hint: Sodium carbonate $N{a_2}C{O_3} \cdot 10{H_2}O$ , is a inorganic compound with the formula $N{a_2}C{O_3}$ and its various hydrates. It is also known as washing soda, soda ash, and soda crystals. All of the forms are water-soluble, white, odourless salts that produce moderately alkaline solutions in water.

Oxygen is the chemical element with the atomic number $8$ and the symbol $O$ . It is a highly reactive nonmetal and an oxidising agent that readily forms oxides with most elements and other compounds. It belongs to the chalcogen group in the periodic table.
Here, we can find that $N{a_2}C{O_3} \cdot 10{H_2}O$ contains $13$ moles of oxygen, $O$ atoms.
That is, the number of moles of $O$ atoms in $1$ mole of $N{a_2}C{O_3} \cdot 10{H_2}O = 13$
So, the number of moles of $O$ atoms in $0.2$ mole of $N{a_2}C{O_3} \cdot 10{H_2}O = 13 \times 0.2 = 2.6$
It is familiar that, $1$ mole of a substance contains $6.022 \times {10^{23}}$ particles of the substance.
Therefore, the number of moles of $O$ atoms in $2.6$ mole of oxygen $= 2.6 \times 6.022 \times {10^{23}} = 15.66 \times {10^{23}}$
Hence, there are $15.66 \times {10^{23}}$ atoms of oxygen in $0.2$ mole of $N{a_2}C{O_3} \cdot 10{H_2}O$
So, the answer is $15.66 \times {10^{23}}$ atoms.