Question

What volume in mL of 0.300 M NaF would be required to make a 0.0715 M solution of NaF when diluted to 250.0 mL with water

Answer 1

Answer: The volume of 0.300 M NaF required is 59.6 ml

Explanation:

According to the dilution law,

$C_1V_1=C_2V_2$

where,

$C_1$ = concentration of concentrated NaF solution = 0.300 M

$V_1$ = volume of pure concentrated Na F solution = ?

$C_2$ = concentration of diluted NaF solution= 0.0715 M

$V_2$ = volume of diluted NaF solution= 250.0 ml

$0.300\times x=0.0715\times 250.0$

$x=59.6ml$

Thus the volume of 0.300 M NaF required is 59.6 ml