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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 1mole1\,mole of Na3PO4?N{a_3}P{O_4}\,?
(i) 4.72×1024atoms4.72 \times {10^{24}}\,atoms
(ii) 4.82×1024atoms4.82 \times {10^{24}}\,atoms
(iii) 4.82×1023atoms4.82 \times {10^{23}}\,atoms
(iv) 4.72×1023atoms4.72 \times {10^{23}}\,atoms

Explanation

Solution

For calculating the number of atoms present in 1mole1\,mole of Na3PO4N{a_3}P{O_4} you must use the fact that 1mole1\,mole of any substance contains 6.022×10236.022 \times {10^{23}}atoms, molecules or formula units of the substance. Since Na3PO4N{a_3}P{O_4} is an ionic compound 1mole1\,mole of Na3PO4N{a_3}P{O_4} will contain 6.022×10236.022 \times {10^{23}} formula units of Na3PO4N{a_3}P{O_4}. Calculate the number of atoms present per formula unit of Na3PO4N{a_3}P{O_4} 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)\left( {6.022 \times {{10}^{23}}} \right) of atoms, molecules or formula units of the substance.
Since Na3PO4N{a_3}P{O_4} is an ionic compound, 1mole1\,mole of Na3PO4N{a_3}P{O_4} will contain 6.022×10236.022 \times {10^{23}} formula units of Na3PO4N{a_3}P{O_4}.
Now we need to calculate the number of atoms present in one formula unit of Na3PO4N{a_3}P{O_4}.
The chemical formula is Na3PO4N{a_3}P{O_4} hence one formula unit contains 33 atoms of NaNa, 11 atom of PP and 44atoms of OO.
Therefore the total number of atoms in one formula unit of Na3PO4N{a_3}P{O_4} =( = \,\,(\,Number of atoms of Na)+(Na\,) + \,(\,Number of atoms of P)+(P\,) + \,(\,Number of atoms of O)O\,)
=(3+1+4)=8= \,\left( {3 + 1 + 4} \right)\,\, = \,\,8
Hence one formula unit of Na3PO4N{a_3}P{O_4} contains 88 atoms.
Now, one mole of Na3PO4N{a_3}P{O_4} contains 6.022×10236.022 \times {10^{23}} formula units of Na3PO4N{a_3}P{O_4}.
\Rightarrow One mole of Na3PO4N{a_3}P{O_4} will contain 6.022×1023formulaunits×8atoms1formulaunit6.022 \times {10^{23}}\,formula\,\,units\, \times \,\dfrac{{8\,atoms}}{{1\,formula\,\,unit}} of Na3PO4N{a_3}P{O_4}.
\Rightarrow One mole of Na3PO4N{a_3}P{O_4} will contain 4.8176×1024atoms4.8176 \times {10^{24\,}}\,atoms of Na3PO4N{a_3}P{O_4}.
So one mole of Na3PO4N{a_3}P{O_4} will contain 4.82×1024atoms4.82 \times {10^{24}}\,atomsof Na3PO4N{a_3}P{O_4}.

Hence the correct answer is (ii) 4.82×1024atoms4.82 \times {10^{24}}\,atoms.

Note: We must remember a mole of a substance always contains 6.022×10236.022 \times {10^{23}} 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.