Question

How many $N{{a}^{+}}$ ions are present in 100ml of 0.25M NaCl solution?
[A] $0.25\times {{10}^{23}}$
[B] $1.505\times {{10}^{22}}$
[C] $15\times {{10}^{22}}$
[D] $2.5\times {{10}^{23}}$

Answer 1

Hint: To find the number of sodium ions in the solution, firstly use the molarity to find the number of moles in the solution. Multiplying it by the Avogadro’s number will give you the amount of sodium cations present.

Complete step by step answer:
 Here, the concentration of the solution is given to us in terms of molarity. We know that molarity is the number of moles of solute per litre of solvent. Therefore, we can use the formula of molarity to find out the number of moles of solute.
Also, the volume is given to us in mL which we have to convert in litre.
We know that 1 litre is 1000 millilitre. Therefore, we can write that 1mL is one-thousandth of a litre. The volume of solution given in the question is 100mL which we can convert into litres using the above information.

Now, let us calculate the number of moles of solute present in the solution.
We know,
$\begin{align}
  & molarity=\dfrac{number\text{ of moles of solute}}{\text{volume of solvent(litre)}} \\
 & or,number\text{ of moles of solute}=molarity\times volume\text{ of solvent} \\
\end{align}$
Or, number of moles of solute = 0.25M $\times $100mL = 0.25M$\times \dfrac{100}{1000}L$
Therefore, number of moles of solute = 0.025 mol
Now, we know that Avogadro’s number is the number of units of one mole of any substance.
Therefore, to find the number of ions of sodium, multiplying the number of moles by Avogadro’s number will give us the answer.

We denote Avogadro’s number by ${{N}_{A}}$ and its value is $6.023\times {{10}^{23}}$.
So, the number of ions present = $0.025\times 6.023\times {{10}^{23}}=1.505\times {{10}^{22}}$.
Therefore, number of sodium cations present in 100mL of 0.25M NaCl solution is $1.505\times {{10}^{22}}$
So, the correct answer is “Option B”.

Note: We should not be confused between molarity, normality and molality. Even though all of them are concentration terms, they are not the same. Normality is the gram equivalents of solute present per litre of solvent whereas molality is number of solute present per kilogram of the solvent and we have already discussed molarity in the above question i.e. number of moles of solute per litre of solvent.