Okay this is a very long question; sorry. 2.0184 g of a compound of the general

Question

Okay this is a very long question; sorry.

2.0184 g of a compound of the general formula Crx(NH3)yClz was prepared using 3.5246 g CrCl3 * 6H2O and 10.00 mL of 15 M NH3. It was analayzed as follows:

Analysis of Cr+3

The concentration of Cr+3 was determined spectrophotometrically. The chromium (III) ion is purple, with a maxiumum light absorption of 570 nm. A stock solution (solution 1) of 0.100 M Cr+3 was used to make the following dilutions:

> Solution 2: 10.00 mL stock solution diluted to a total of 100 mL with water.

> Solution 3: 20.00 mL stock solution diluted to a total of 100 mL with water.

1.2532 g of the unknown chromium compound was dissolved in enough water to make 100 mL of solution. Absorbance readings are given below:

Calculate the concentration of Cr+3 in solutions #2 and #3. Then make a graph of absorbance vs. concentration using the three known (standard) solutions. From the graph, find the concentration of the chromium compound, calculate moles of Cr+3, grams of Cr+3, and % Cr+3 in the unknown compound. Attach the graph and show your work.

Explanation / Answer

So we have here,

Stock solution of Cr3+ = 0.1 M .....soln. #1

0.1 M x 10 ml/100 ml = 0.01 M ......soln. #2

0.1 M x 20 ml/100 ml = 0.02 M ......soln. #3

Prepare a calibration curve

x-axis : conc. (M)

y-axis : abs. (nm)

We get the equation,

y = -0.8315x + 0.24667

for unknown we have an absorbance of = 1.22

1.22 = -0.8315x + 2.4667

x = 1.50/100 = 0.015 M

So the concentration of Cr3+ in unknown solution is 0.015 M

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