Question

You prepare a 2nd solution (solution B) by pipetting 10.00 ml of your 1st solution (solution A) into a 50.00 ml volumetric flask and diluting with deionized water to the 50.00 ml mark.
If the concentration of solution B is 0.1045 M NaCl, what was the NaCl concentration of solution A?

Answer 1

Answer:

Concentration solution A was 0.5225 M

Explanation:

10.00 mL of solution A was diluted to 50.00 mL and yields 50.00 mL of solution B

According to laws of dilution-      $C_{A}V_{A}=C_{B}V_{B}$

where, $C_{A}$ and $C_{B}$ are concentration of solution A and B respectively

$V_{A}$ and $V_{B}$ are volumes of solution A and B respectively

Here $C_{B}$ = 0.1045 M, $V_{B}$ = 50.00 mL and $V_{A}$ = 10.00 mL

Hence, $C_{A}=\frac{C_{B}V_{B}}{V_{A}}=\frac{(0.1045M\times 50.00mL)}{10.00mL}=0.5225M$

So, concentration solution A was 0.5225 M