In a gas chromatography experiment, a mixture of two compounds (A and B) was separated for analysis. A chart recorder that fed paper at a constant rate of 100 mm. per minute was used to record the chromatogram, and a mark was made on the chart at the time the sample was injected. The first peak on the chromatogram corresponds to compound A, and that peak was observed at a distance of 5.83 cm from the sample injection point, and it had a height of 3.52 cm and a width at half-height of 1.25 cm. The peak corresponding to B was observed at a distance of 7.96 cm from the injection point, and it had a height of 3.34 cm and a width at half-height of 1.06 cm. What was the retention time for compound B (in seconds)? (Enter your answer as a number only.)

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valetta valetta · Answer 1 · 2019-10-10T12:02:05+02:00

Answer:

47.76 seconds

Explanation:

Given:

Rate = 100 mm per minute

Distance of compound A from the Injection point = 5.83 cm = 58.3 mm

Distance of compound B from the Injection point = 7.96 cm = 79.6 mm

Now,

The retention time = [tex]\frac{\textup{Distance}}{\textup{Rate}}[/tex]

thus,

For the compound B

The retention time = [tex]\frac{\textup{79.6 mm}}{\textup{100 mm/minute}}[/tex]

or

The retention time = 0.796 minute

also,

1 minute = 60 seconds

therefore,

0.796 minutes = 0.796 × 60 seconds = 47.76 seconds