Question
Question: How many atoms are present in \(1\,mole\) of \(N{a_3}P{O_4}\,?\) (i) \(4.72 \times {10^{24}}\,atom...
How many atoms are present in 1mole of Na3PO4?
(i) 4.72×1024atoms
(ii) 4.82×1024atoms
(iii) 4.82×1023atoms
(iv) 4.72×1023atoms
Solution
For calculating the number of atoms present in 1mole of Na3PO4 you must use the fact that 1mole of any substance contains 6.022×1023atoms, molecules or formula units of the substance. Since Na3PO4 is an ionic compound 1mole of Na3PO4 will contain 6.022×1023 formula units of Na3PO4. Calculate the number of atoms present per formula unit of Na3PO4 and hence calculate the number of atoms present in one mole.
Complete solution:
We know, per mole of any substance contains Avogadro's number (6.022×1023) of atoms, molecules or formula units of the substance.
Since Na3PO4 is an ionic compound, 1mole of Na3PO4 will contain 6.022×1023 formula units of Na3PO4.
Now we need to calculate the number of atoms present in one formula unit of Na3PO4.
The chemical formula is Na3PO4 hence one formula unit contains 3 atoms of Na, 1 atom of P and 4atoms of O.
Therefore the total number of atoms in one formula unit of Na3PO4 =(Number of atoms of Na)+(Number of atoms of P)+(Number of atoms of O)
=(3+1+4)=8
Hence one formula unit of Na3PO4 contains 8 atoms.
Now, one mole of Na3PO4 contains 6.022×1023 formula units of Na3PO4.
⇒ One mole of Na3PO4 will contain 6.022×1023formulaunits×1formulaunit8atoms of Na3PO4.
⇒ One mole of Na3PO4 will contain 4.8176×1024atoms of Na3PO4.
So one mole of Na3PO4 will contain 4.82×1024atomsof Na3PO4.
Hence the correct answer is (ii) 4.82×1024atoms.
Note: We must remember a mole of a substance always contains 6.022×1023 particles called the Avogadro’s number. Also do not forget that the given molecule is ionic hence we used the concept of formula units. Make sure to calculate the number of atoms present in one formula unit only.