Question

To what volume (in milliliters) should you dilute 100.0 mL of a 5.50 M solution of CaCl2 solution to obtain a 0.950 M solution of CaCl2?

Answer 1

Dilution law is given as:

$M_{1}V_{1}=M_{2}V_{2}$    (1)

where, $M_{1}$ = molarity of initial concentrated solution.

$V_{1}$ = volume of initial concentrated solution.

$M_{2}$ = molarity of final diluted solution.

tex]V_{2}[/tex] = volume of final diluted solution.

Volume of initial concentrated solution of $CaCl_{2}$= 100.0 mL

Molarity of initial concentrated solution of $CaCl_{2}$= 5.50 M

Volume of final diluted solution of $CaCl_{2}$ = 0.950 M

Put the values in formula (1),

$5.50 M\times 100.0 mL=0.950 M\times V_{2}$

$V_{2} =\frac{5.50 M\times 100.0 mL}{0.950 M}$

$V_{2} =578.9 mL$

Hence, final diluted volume of the solution is  $578.9 mL$.