Discussion
A compound elutes at \( 300 \ \mathrm{s} \) with a dead time of \( 60 \ \mathrm{s} \). Its capacity factor is
Answer: D
Subtracting the dead time before dividing:
\[ k = \frac{300-60}{60} = 4 \]
giving 4, meaning the compound spends four times as long in the stationary phase as in the mobile. Five would follow from forgetting to subtract, and 240 is the adjusted time in seconds. A value between two and ten is usually the target, since below that resolution suffers and above it the analysis simply takes longer.
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