[redfox]

Ok, here's my question:

A solution containing a mixture of 300mg of A (RMM=90.25) dissolved in 30cm³ of solvent, and an unknown quantity of B (RMM = 120.1) dissolved in 20cm³ of solvent was separated by gas chromatography at 60 degrees C. The relative peak areas of A and B were 1.00:1.20 respectively. Considering A to be the internal standard, calculate the concentration of B in moles/Litre, if the response factor, F, was 1.065.
Briefly explain the significance of the response factor value, F, that was used in the calculation.

The equation is Ax/[ x ]=F As/[ s ]
As (internal standard) is 1.00
Ax (analyte) is 1.20

[ s ] I think is 300mg/90.25...which would be .3g/90.25 = 3.32x10^-3/.03dm³=0.11mol dm³

[ x ] is unknown

I did make a start on this but it seemed silly as I don't know if [ x ] should just be x til I figure it out, or if I should put 1/(x/120.1)

Any help offered is much appreciated! Thanks:)

[Mr Peanut]

OK, first part first.

We are mixing the 30 ml and 20 ml solutions so the final volume is 0.05 liters.
Cis = 300/0.05 mg/l.

If (and this is an important if) your text defines RF= (AsCis)/(AisCs) then

Cs=(1.2Cis)/RF=...

Your text may define RF differently. (There is no universal definition.) Let me know.