Question

Assume that you carry out the following dilutions:
(i) you take 1mL of 10^-6M of AsCl3 and diluted it to 1 Kiloleter (Solution A)
(ii) Then you take 1microliter of Solution A and dilute to 10L to make solution B
(iii) Then you take 1mL of Solution B and dilute to 1L to make Solution C.

A) Calculate the concentration of AsCl3 in Solution C [I got an answer for A) which is 1.0 * 10^-22M (don't know if thats correct)]

B) Calculate the number of molecules of AsCl3 that would be in 1 mL of Solution C at this concentration. Look at your answer. This is the average number per mL. Butwhat is the most likely number (ie integer) of molecules of AsCl3 in 1 mL?

Answer 1

Part A) Calculate the concentration of AsCl₃ in Solution C

Here, we can use the formula C_fV_f = C_iV_i

Cf is the final concentration

Vf is the final volume

Ci is the initial concentration

Vi is the initial volume

(i) You take 1mL of 10^-6 M of AsCl₃ and diluted it to 1 Kiloliter (solution A)

[(1x10^-3 L)(10^-6mol/L)] / (1 x 10³) = 1 x 10^-12 mol/L

Concentration of solution A = 1 x 10^-12 M

(ii) Then you take 1microliter of Solution A and dilute to 10 L to make solution B

[(1 x10^-6 L)(1 x 10^-12 mol/L)] / (10 L)= 1 x 10^-19 mol/L

Concentration of solution B = 1 x 10^-19 M

(iii) Then you take 1mL of Solution B and dilute to 1L to make Solution C.

(1x10^-3L)(1 x 10^-19 mol/L) / 1L = 1 x 10^-22 mol/L

Concentration of solution C = 1 x 10^-22 M

Part B) Calculate the number of molecules of AsCl₃ that would be in 1 mL of Solution C

Number of molecules of AsCl3 in 1 mL of solution C = [(1 x 10^-22 M) / (1000 mL)] x 6.022x10²³

                                                                                       = 0.060 molecules