Equal volumes of 0.50 M HCl, 0.25 M NaOH and 0.75 M NaCl are mixed. The molarity of the resultant NaCl solution is

The correct option is B 0.33 M
Let’s take the volume as V L of each solution.
Moles of HCl, NaOH and NaCl present initially = 0.50V, 0.25V and 0.75V respectively.
$HCl+NaOH\to NaCl$
0.50V 0.25V - $\text{(number of moles before reaction)}$
HCl and NaOH react to produce 0.25V M of NaCl
Total moles of NaCl in the solution = 0.25V (after reaction) + 0.75V (given)
Total moles of NaCl in the solution = V moles
$\text{Molarity}=\frac{\text{total moles of NaCl}}{\text{total volume in L}}=\frac{V}{3V}MM=\frac{1}{3}M=0.33M$