A. To determine the rate of reaction for the decomposition of aspirin a student

Question

A. To determine the rate of reaction for the decomposition of aspirin a student weighted out 0.0650 g of aspirin (MW 180.57 g/mol) and transferred it to a test tube. In the test tube he added 1.00 mL of n- and 10.00 ml water. The test tube with the mixture aspirin was first heated in the water bath at 92 ec for 6 and after removed from the bath and cooled down. To determine minutes the amount of salicylic acid obtained in the aspirin degradation, first, 5.00 ml of 0.025 M Fe(NO,) was added to obtain the violet complex of iron-salicylate and second the absorbance of the violet solution was measured at 525 nm with a S-20 spectrometer and found to be 0.304. a) What is the molarity of the aspirin solution prepared, before being heated to 92 °C? Concentration of aspirin in test tube before reaction mol/L. B. Standard stock solution of salicylic acid is prepared by dissolving 0.0799 g of sali c acid (Mw 138.12 g/mol in water and diluting to 50.00 mL. AS a) what is the concentration of salicylic acid in the stock solution express in mol/L? Concentration of salicylic acid in stock solution O mol/L, C. The following standard salicylic acid solutions (2.31 x 10 4.63 x 10 6.94 x10 9.26 x 10 and 11.6 10 mol/LI were prepared from the stock solution (each standard solution was prepared in an amount of 10.00 mLI. The absorbance of the standard solutions is obtained from the S-20 spectrometer at the wavelength of 525 nm as shown in the data table below: Absorbance Volume of Stock Concentration of salicylic acid Entry solution used (mu) standard solution (mol/L) 0.070 Standard Solution 1 Standard Solution 2 2Y uL 4.63 x 104 0.150 Standard Solution 3 LuL 6.94 x 10 u 9.26 x 104 0.250 Standard Solution 4 0.3600 Standard solution 5

Explanation / Answer

Question 1

a) Molarity of the aspirin solution.

Let's gather the data we have:

mass of aspirine = 0.0650g (be careful because on the calculations you show, you used 0.650g).

molar mass of aspirine = 180.57g/mol

volume of solution = 10mL water + 1mL n-propanol = 11mL*

*It is very important to make an observation here: the volume we use for molarity is the volume of the solution, not just the volume of solvent (water in this case); volumes generally don't add up, it is possible the final volume of solution is 10.5mL or 10.7mL or it can be 11mL; this is because the molecules of the liquids here intereact with each other, since we don't have any further information about these interactions, we will consider that the final volume of this solution is 11mL indeed*

So, to calculate M we need to follow the formula:

M = moles aspirin/L solution

And to calculate moles we use the formula:

moles aspirin = grams/molar mass

So our M formula becomes:

M = (grams/molar mass) / L solution

M = (0.0650g/180.57g/mol) / 0.011L

M = 0.0327 mol/L

b) We dissolve m = 0.0799g of salicylic acid (molar mass = 138.12g/mol) in 50mL of water.

We use the same formula from above to calculate M

M = (grams/molar mass) / L solution

M = (0.0799g/138.12g/mol) / 0.05L

M = 0.0116 mol/L

c) To complete the table we need to know how much of the above solution we used to make the dilutions, we will use the formula:

C1V1 = C2V2

(0.0116M)*V1 = C2*(10.mL)

Where V1 is the volume of the stock solution we need (the second column in the table) and C2 is the concentration of the desired standard solution (the third column in the table).

So our table looks like this:

For the first standard:

(0.0116M)*V1 = C2*(10mL)

(0.0116M)*V1 = (0.000231M)*(10mL)

V1 = 0.199mL

V1 = 199uL

Repeat the procedure for the rest of the solutions and the table will look like this:

Stock volume (V1) (uL = microliters) Standard Concentration (C2) (mol/L) 199 0.000231 0.000463 0.000694 0.000926 0.001160

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