You have an unknown dye solution whose concentration seems to lie between a 0.00

Question

You have an unknown dye solution whose concentration seems to lie between a 0.0006 M stock solution and a 0.0004 M stock solution. Therefore, you had planned to dilute the 0.0006 M stock to create your calibration standards. However, you see that there is no 0.0006 M stock left, but there is plenty of 0.0004 M stock.

a. How could you raise the concentration of some of the 0.0004 M stock to make 75 mL of a 0.0006 M solution without adding any more dye? Be sure to show calculations and indicate the volume of 0.0004 M required.

b. Draw a molecular view picture illustrating the moles of dye remaining constant during this process.

Explanation / Answer

a) If we take 75 ml of 0.0006 M solution, using the relation Conc = n / V ,
the no. of moles of dye in this solution will be , n = C x V = 0.075 x 0.0006 = 4.5 x 10^-5 moles .

Now to get 4.5 x 10^-5 moles of dye of a 0.0004 M solution, again using C= n / V so,V =n/C

112.5 ml of 0.0004 M solution is required.

b) Now what you need to do is to remove 112.5 - 75 = 37.5 ml of solution from the 112.5 ml of 0.0004 mol solution, however only the pure water (no dye dissolved) must be removed, for this if the dye settles down in the water after some time of sedimentation , the pure water of required volume can be removed by using a pipette to accurately remove 37.5 ml of solution, if the dye is completely soluble, then we have a much difficult problem , this time distillation can be used to remove the water only, which will obivously be complicated

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