QUESTION 4 You have a S mg/ml solution of caffeine (molar mass O 12.9 uM O 19.4

Question

QUESTION 4 You have a S mg/ml solution of caffeine (molar mass O 12.9 uM O 19.4 uM O 129 uM O 194.2 UM 194.2 g/mol), You diluted that solution 200 times (200 fold]). Calculate the final concentration (in solution 200 times (2 QUESTION5 Upon injecting 3040 units, r equation (Trendline, intercept set to zero) You then 20 ul aliquots of 1, 2, 4,8, and 16 uM construct a calitbration with an unknown amount of compound X That sample gave a peak with area of 1900. What was the of pure compound X into HPLC column, you got peak areas of 190, 420, 780, 1600, and plot: peak area against concentration of compound X (nM), and fit the plot with a linear (NB: area can be plotted against either concentration or amount. For example, 20 ul of 16 uM solution corresponds t you inject the same volume for both the concentration of the analyte in your sample) O 9.8 nM O 9.8 UM O 190 uM O 32.0 UM standards and the sample, it is convenient to use concentration -after all, what you need at the end is the of o 320 pmol of compound ?. if

Explanation / Answer

Ans 4 : 129uM

Let the amount of solution be 1 mL

Number of moles of caffeine = 0.005 / 194.2 = 0.0000257 mol

So when the dilution is 200 fold , the volume of solution becomes 200 mL

Molarity = no. of moles / volume of solution in L

= 0.0000257 / 0.200

= 0.0001285 M

= 129uM

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