Question

Assume that you did the following dilutions

1) You take 1mL of 10^-6 M of AsCl3 and diluted it to 1 kilometer (solution A)
2) you the take 1 micrometer of solution A and dilute 10L to make solution B
3)then you take one mL of solution B and dilute to 1L to make solution C
Find the concentration of AsCl3 in solution C

Answer 1

Below is the solution. I hope it helps.

CfVf = CiVi
Cf = (CiVi)/ Vf

i. Cf = [ (10^-6 mol / L) (1 mL) (1L / 1000 mL) ] / [ (1kL) (1000L / 1 kL) ] = 1x10^-12 M → use as Ci in next dilution
ii. Cf = 1x10^-19 M → use as Ci in next dilution
iii. Cf = 1x10^-22 M

Answer 2

The concentration of AsCl3 in solution C is
`10^(-22)M`

Step-by-step explanation:

To solve this problem we need the next equation

(final concentration)(final volume) = (initial concentration)(initial volume) or

(Cf)(Vf)=(Ci)(Vi)

and we are going to work in liters, so be aware of the units

For all the steps we need to find the final concentration, Cf, so:

`Cf=((Ci)(Vi))/(Vf)`

1)
`Cf=((10^(-6) M )(0.001L))/(1000L) =10^(-12)M`

Note that 1mL = 0.001L and 1kL = 1000L

[A] =
`10^(-12)M`

2) now Ci = [A] so:

`Cf=((10^(-12) M )(10^(-6)L))/(10L) =10^(-19)M`

Note that 1uL =
`10^(-6)L`

[B] =
`10^(-19)M`

3) now Ci = [B] so:

`Cf=((10^(-19) M )(0.001L))/(1L) =10^(-22)M`

[C] =
`10^(-22)M`