Question

To what volume should you dilute 319 mL of a 6.70 M calcium nitrate so that 44.5 mL of the diluted solution contains 1.972 g calcium nitrate?

Answer 1

mass of Ca(NO3)2 = 1.972 g

moles of Ca(NO3)2 = 1.972 / 164.09 = 0.0120 mol

volume = 44.5 mL = 0.0445 L

concentration of diluted solution = 0.0120 / 0.0445

                                                   = 0.270 M

volume V1 = 319 mL

concentration C1 = 6.70

concentration C2 = 0.270 M

volume V2 = ??

C1 V1 = C2 V2

6.70 x 319 = 0.270 x V2

V2 = 7916 mL

volume of diluted solution = 7916 mL