Answer / kangchuentat

Answer 95 - Let (f-factor) = (analyte signal) (concentration of internal standard) / [ (internal standard signal) (concentration of analyte) ]. For first solution, (f-factor) = (2 mamps) (4 mM) / [ (3 mamps) (3 mM) ] = 8 / 9. For second solution, (f-factor) = (1 mamps) (1 mM) / [ (4 mamps) (2 mM) ] = 1 / 8. Average : Average f-factor = (8 / 9 + 1 / 8) / 2 = (64 + 9) / (72 x 2) = 73 / 144. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question 96 - In infrared spectrum, one of the factors affecting peak location is the mass of the atoms. The stretching frequency of a bond connected to a lighter atom will be greater than the same bond connected to a heavier atom. (a) For halogens like florine (F), chlorine (Cl), bromine (Br), iodine (I) and astatine (At), what is their IUPAC group number? Hint : The proton numbers for F, Cl, Br, I and At are 9, 17, 35, 53 and 85 respectively. (b) For the compounds of H-F, H-Cl, H-Br, H-I and H-At, which one has the lowest stretching frequency and which one has the highest stretching frequency? State the reasons.

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Question 95 - The following formula is used as a reference : (analyte signal) / (internal standard signal) = (f-factor) x (concentration of analyte) / (concentration of internal standard). A solution containing 3 mM of analyte and 4 mM of internal standard gave peak signals of 2 and 3 mamps respectively. Another similar solution containing 2 mM of analyte and 1 mM of internal standard gave peak signals of 1 and 4 mamps respectively. Find the average f-factor.

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