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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

Verified Answer :

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 

mppalenciaa

Answer:

The concentration of AsCl3 in solution C is  \(10^{-22}M\)

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:

\(Cf=\frac{(Ci)(Vi)}{Vf}\)

1) \(Cf=\frac{(10^{-6} M )(0.001L)}{1000L} =10^{-12}M\)

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

   [A] = \(10^{-12}M\)

2) now Ci = [A] so:

   \(Cf=\frac{(10^{-12} M )(10^{-6}L)}{10L} =10^{-19}M\)

   Note that 1uL = \(10^{-6}L\)

   [B] = \(10^{-19}M\)

3) now Ci = [B] so:

   \(Cf=\frac{(10^{-19} M )(0.001L)}{1L} =10^{-22}M\)

   [C] = \(10^{-22}M\)

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