For an unknown using the retention times 1. 6 min for ch4, 12. 8 min for octane, 15. 4 min for the unknown, and 19. 6 min for nonane, the kovats retention index is mathematically given as
I = 807.857
Generally, the equation for the Kovats index(i) is mathematically given as
[tex]I= \frac{100[n + (N - n) x (Log tr (unknown) - logtr (n))}{ logtr(N) - logtr(n))]}[/tex]
Therefore
[tex]I= \frac{100[ 1+8*(log(14.7) -log (1.5)}{log (19.8) - log (1.5)) 0]}[/tex]
I = 100[ 1+8*(0.991/1.12)]
I = 807.857
In conclusion, Kovats retention index
I = 807.857
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Kovats retention index is a system that converts the retention time into the independent constants of the system. The Kovats retention index for the unknown is 807.857.
The Kovats retention index (I) is the system used in analytical techniques like gas chromatography for the conversion of the retention time. It is given mathematically as:
[tex]\rm I = \rm \dfrac{100[n+(N-n)x(Logtr(unknown)-logtr(n))}{Logtr(N)-Logtr(n)]}[/tex]
Substituting values and solving further,
[tex]\begin{aligned} \rm I &= \rm \dfrac{100[1+8 \times (Log(14.7) - log (1.5))}{(log (19.8) - log (1.5))0}\\\\&= 100 [ 1 + 8 (\dfrac{0.991}{1.12})]\\\\&= 807.857\end{aligned}[/tex]
Therefore, the Kovats retention index for an unknown is 807.857.
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