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Question: 5.85 g NaCl is dissolved in 1 L water. The number of ions of \({ Na }^{ + }\) and \({ Cl }^{ - }\) i...

5.85 g NaCl is dissolved in 1 L water. The number of ions of Na+{ Na }^{ + } and Cl−{ Cl }^{ - } in 1 mL of this solution will be:
(a) 6.02×1019\times { 10 }^{ 19 }
(b) 1.2×1022\times { 10 }^{ 22 }
(c) 1.2×1020\times { 10 }^{ 20 }
(d) 6.02×1020\times { 10 }^{ 20 }

Explanation

Solution

NaCl is a strong electrolyte and hence it dissociates completely in water. Therefore one formula unit of NaCl will dissociate to give one Na+{ Na }^{ + } ion and one Cl−{ Cl }^{ - } ion, a total of two ions.

Complete step by step answer:
We can solve this question in the following way:
The molar mass of NaCl is 58.44 g/mol. The number of moles of a substance is the ratio of is mass to its molar mass:
Number of moles of a substance (n)=mass in gMolar mass in g/mol\begin{matrix} Number\ of\ moles \\\ of\ a\ substance\ (n) \end{matrix} = \cfrac { mass\ in\ g }{ Molar\ mass\ in\ g/mol }
⇒Number of moles of a NaCl (n)=5.85 g58.44 g/mol=0.10 mol\Rightarrow \begin{matrix} Number\ of\ moles \\\ of\ a\ NaCl\ (n) \end{matrix} = \cfrac { 5.85\ g }{ 58.44\ g/mol } = 0.10\ mol.

0.10 mole of NaCl is present in 1 Litre of water, Hence the number of moles of NaCl present in 1 mL of the solution is = 0.1. mol1000 mL×1 mL=0.0001 mol\cfrac { 0.1.\ mol }{ 1000\ mL } \times 1\ mL = 0.0001\ mol.

NaCl is a strong electrolyte, therefore it will dissociate completely into its constituent ions. Therefore one formula unit of NaCl will dissociate to give one Na+{ Na }^{ + } ion and one Cl−{ Cl }^{ - } ion; in total two ions. Therefore 0.0001 moles of NaCl will give 0.0001 moles of Na+{ Na }^{ + } ions and 0.0001 moles of Cl−{ Cl }^{ - } ion. Hence the total number of ions in 0.0001 moles of NaCl is 0.0002 moles.

Now, 1 mole of a species contains 6.023×10236.023\times { 10 }^{ 23 } number of atoms/molecules/ions/any other species. Therefore 0.0002 moles will have = 0.0002×6.023×1023ions=1.205×1020ions0.0002\times 6.023\times { 10 }^{ 23 }ions = 1.205\times { 10 }^{ 20 }ions .

Hence the correct answer to the above question is (c) 1.2×1020\times { 10 }^{ 20 }.

Note: Always remember that in this case the electrolyte given was a strong electrolyte, therefore it dissociated completely. This is not the case with weak electrolytes which only partially dissociate in a solution because of their low value of the dissociation constant. In such a case if we want to know the number of ions formed by the weak electrolyte, then we would have to calculate its degree of dissociation.