Dilution Problem. No one on Chegg seems to be able to answer it correctly. Pleas

Question

Dilution Problem. No one on Chegg seems to be able to answer it correctly. Please help.

You are wording in the laboratory and are given a 1 0 times 10^-2 M erythrosine dye solution You are then asked to make 100.0 mL of a 3.2 times 10^-6 M erythrosine dye solution. Describe with words and calculations how you make this solution using only glassware and equipment that was available in the laboratory during the semester Describe how you would make 10 0 mL of a 4 0 times 10^-4 M solution of KSCN from using reagent grade, powdered KSCN Note that your experimental procedure should produce minimal errors and that only glassware and equipment used during the semester can be used

Explanation / Answer

Dilution

(i) To prepare 100 ml of 3.2 x 10^-6 M solution starting with stock 1.0 x 10^-2 M solution

Using,

C1V1 = C2V2

with,

C1 nd C2 be the concentration of stock and final solution, respectively.

V1 and V2 be the volumes of stock and final solution, respectively.

So,

V1 = 3.2 x 10^-6 M x 100 ml/1.0 x 10^-2 M

      = 0.032 ml

So you take 32 microliter of 1.0 x 10^-2 M solution and dilute it to 100 ml to get a 3.2 x 10^-6 M solution of erythrosine.

(ii) To prepare a 10 ml of 4.0 x 10^-4 M solution of KSCN

molarity = grams of solute/molar mass of solute x volume of solution (L)

So,

grams of powdered KSCN required = 4.0 x 10^-4 M x 0.010 L x 97.2 g/mol

                                                         = 0.0004 g

                                                         = 0.4 mg

So, we take 0.4 mg of powdered KSCN and dissolve it in 10 ml of water to get 10 ml of 4.0 x 10^-4 M KSCN solution.

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