Question

10 mL of a stock solution of 1 M HCl is serially diluted thrice. Each step of the serial dilution dilutes the original solution by half. What will be its final concentration?

• A. 0.5 M
• B. 0.125 M
• C. 0.025 M
• D. 0.25 M

Answer 1

The correct option is B 0.125 M

For first dilution
$M_1{V}_1=M_2{V}_2$
where $M_1$ and $V_1$ are the molarity and the volume of the initial solution respectively.
$M_2$ and $V_2$ are the molarity and the volume after first dilution.
$\frac{M_1{V}_1}{V_2}=M_2$
So, $M_2$ = 0.5 M
For second dilution,
$M_2{V}_2=M_3{V}_3$
where ,$M_3$ and $V_3$ are the molarity and the volume after second dilution.
$\frac{M_2{V}_2}{V_3}=M_3$
So, $M_3$ = 0.25 M
For third dilution,
$M_3{V}_3=M_f{V}_f$
where, $M_f$ and $V_f$ are the molarity and the volume of the final solution.
$\frac{M_3{V}_3}{V_f}=M_f$
So, $M_f$ = 0.125 M
On each dilution, the concentration is halved, and the total volume is doubled.