A standard solution containing 6.3 × 10-8 M iod°acetone and 2.0 x 10-7 M p- dich

Question

A standard solution containing 6.3 × 10-8 M iod°acetone and 2.0 x 10-7 M p- dichlorobenzene (an internal standard) gave peak areas of 395 and 787, respectively, in a gas chromatogram. A 3.00 mL unknown solution of iodoacetone was treated with 0.100 mL of 1.6 × 10-5 M p-dichlorobenzene and the mixture was diluted to 10.00 mL. Gas chromatography gave peak areas of 633 and 520 for iodoacetone and p-dichlorobenzene, respectively. Find the concentration of iodoacetone in the 3.00 mL of original unknown.

Explanation / Answer

The ratio of concentration of iodooctane and p-dichloro benzene is (6.3*10-8 : 2.0 * 10-7) = 0.315

The ratio of peak areas obtained for iodooctane and p-dichloro benzene is 395/ 787 = 0.502

Suppose the concentration of iodooctane in 10 mL of diluted solution is x M.

As internal standard 0.100 mL of 1.6 * 10 -5 M p- dichloro benzene has been used.

In 10 mL. of diluted solution, concentration of p- dichloro benzene is = 0.1 * 1.6 * 10 -5/ 10 M = 1.6 * 10-7 M

The ratio of peak areas of iodo octane and p- dichloro benzene = (633/520) = 1.217

When, peak ratio is 0.502, concentration ratio is 0.315

So, when peak ratio is 1.217, concentration ratio is (0.315 * 1.217)/ 0.502 = 0.764

So, x / 1.6 * 10-7 = 0.764

x = 1.22 * 10 -7 M

In 10 mL. solution, concentration of iodooctane is 1.22 * 10 -7 M

In 3 mL. (original solution) solution: the concentration of iodooctane is (1.22 * 10 -7 * 10/ 3) M = 4.067 * 10-7 M.

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