1. You have an unknown dye solution whose concentration seems to lie between 0.0

Question

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

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

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

Explanation / Answer

Now if you 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 , you'll need 112.5 ml of 0.0004 M solution

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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