Question

2. The absorbance of a series of standard solutions of a substance X vs concentration [X], in ppm) is measured. The data is as following: Concentration (ppm) 10.0 20.0 30.0 40.0 50.0 Absorbance 0.105 0.198 0.330 0.402 0.533 An unknown solid sample (0.1050 g) containing substance X was dissolved in 5.0 mL acidic solution and then transferred to a 100.0 mL volumetric flask and filled with deionized water to the mark. The absorbance of this solution is 0.378. What is the weight percent of X in the unknown solid sample?

Answer 1

For this question first we need to draw the graph of absorbance of standard solution Vs concentration.

Absorbance Vs Concentration y = 0.0106x -0.0044 R2 = 0.9935 o 10 20 30 40 50 60

The values on left side is for absorbance and values on horizontal axis are for concentrations in ppm. Since the axis title is not available that's why it is mentioned here. The graph will be drawn on excel sheet.

Now, we have equation i.e. y = 0.0106x - 0.0044. Now put the value of unknown absorbance i.e. y in this equation and find the value of x.

Therefore, 0.378 = 0.0106x - 0.0044

Now add both side by 0.0044

0.378 + 0.0044 = 0.0106x

0.3824 = 0.0106x

Divide both side by 0.0106,

(0.3824 / 0.0106) = x

x = 36.0754 ppm

It means there are 36 mg solid sample is present in 1000 ml solution. If we have 100 ml solution then the value of solid sample is

36 mg * (100 ml / 1000 ml) = 3.6 mg sample in 100 ml solution. Now convert this in gram by using dimensional analysis.

3.6 mg * (1g / 1000 mg) = 0.0036 g solid sample

Now find percent sample in unknown sample.

(0.0036 / 0.1050) *100 = 3.4285 % (w/w)

Therefore, percentage of X in unknown solid sample is 3.4 % or 3.43%.

If you find any mistake, please mention in the comment box.

Thanks.