Question

A solution containing 1.50 mM ethanol and 2.40 mM methanol gave peak areas of 1487 and 3544, respectively. Subsequently, 2.00 mL of 3.2 mM methanol was added to a 7.00 mL unknown ethanol and diluted to 10.00 mL. The final solution gave peak areas of 6235 and 4487 for ethanol and methanol, respectively. Calculate the concentration of ethanol in the original unknown.

Answer 1

8.99mM is the concentration of ethanol in the original unknown.

Step-by-step explanation:

In chromatography, the peak areas of compounds are proportional to its concentration. The ratio of Area/Concentration for ethanol and methanol is:

Ethanol = 1487 / 1.50mM = 991,33 mM⁻¹

Methanol = 3544 / 2.40mM = 1476.67mM⁻¹

The concentration of ethanol in the diluted solution is:

6235 / 991.33mM⁻¹ = 6.29mM

The original solution was diluted from 7mL to 10mL, the concentration of the original solution is:

6.29mM * (10mL / 7mL) =

8.99mM