00:01
To answer this question, we need to make an assumption that the first peak corresponds to the dead time, the first peak spending 100 % of its time in the mobile phase, and 0 % of its time in the stationary phase.
00:14
Without this information, we can't answer the question.
00:18
So we'll assume that peak one corresponds to the dead time.
00:23
Thus, all peaks or all compounds spend the same amount of time in the mobile phase in order to travel through the column.
00:30
If they're in the mobile phase, they're moving, and when they're moving, it apparently only takes 0 .85 minutes to get through the column.
00:38
Thus, all peaks spend 0 .85 minutes in the mobile phase.
00:42
The ones that take longer to get through than 0 .85 minutes have additional time spent in the stationary phase.
00:50
Peak 2 spends 2 .75 minus the 0 .85 or 1 .90 minutes in the stationary phase, and peak 3 spends 5 .5 minus .85, or 4 .65 minutes in the stationary phase.
01:07
The resolution between peaks 2 and 3 can be calculated by taking the difference in their retention times divided by the average of their peak width at the base, which corresponds to 4 sigma, given us a resolution of 7 .86, very well separated.
01:26
A chromatogram might look something like this, where the peaks get a little bit broader, as demonstrated by their peak width as the retention time increases...