The following is a description of the preparation of an unknown sample: 1. 0.051

Question

The following is a description of the preparation of an unknown sample: 1. 0.0519g of an unknown Fe2 salt is measured and added to a 25 mL volumetric flask. 1.0 mL of 6 M HCl is added before diluting to volume. 2. 1.0 mL of the solution prepared is transferred to a 100 mL volumetric flask. 1.25 mL of buffer and 5.0 mL of ferrozine is added before diluting to volume with DI water. The absorbance of the solution in 2 is measured to be 0.895. Report all answers to 3 sig figs. The 0.895 absorbance corresponds to 0.06 concentration. calculate the concentration of solution 2? and solution 1? Calculate the number of millimoles of the unknown solution prepared by adding 0.0519 g to 25mL.? and calculate the molar mass of the unknown?

Please provide a step by step solution.

Explanation / Answer

Let the unknown salt be FeX2.

Solution 1 : 0.0519 g of FeX2 + 1 ml 6M HCl + 24 mL water

Solution 2 : 1mL solution 1 + 1.25 mL buffer + 5ml ferrozine + 92.75 mL water

Absorbance , A = 0.895

We know, according to Beer Lambert's law,

Absorbance, A = e*l*c , where

e = molar absorptivity

l = path length

c = molar concentration

This given value corresponds to 0.06M concentration

This means that the concentration of FeX2 in solution 2 is : 0.06 M

Total volume of solution 2 = 100 mL = 0.1 L

Since , molarity = no. of moles / Volume in liters

So, no. of moles of FeX2 in solution 2 = molarity * volume of solution 2 in liters = 0.06*0.1 = 0.006 moles

All these number of moles of FeX2 in solution 2 come from the 1 ml of solution 1 added to it.

This means, 1 ml( = 0.001 L) of solution 1 contains 0.006 moles of FeX2

Thus, molarity of solution 1 = no. of moles/ volume in Liters = 0.006/0.001 = 6M

Thus, concentration of solution 1 = 6M, and

concentration of solution 2 = 0.06 M

Now, since concentration of solution 1 is 6M and total volume of solution 1 is 25 ml( = 0.025L)

So, total number of moles of FeX2 in solution 1 = molarity*volume of solution 1 in liters = 6*0.025 = 0.150 moles

0.150 moles = 150 millimoles

Also, no. of moles = mass added / Molecular weight

Thus, molecular weight MW = mass added / no. of moles = 0.0519/0.150 = 0.346 g

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