Calciferol (Vitamin D) in2.0 g meat of a fatty fish was extracted with hexane and mark up with hexane to 100ml. The %T of calciferol (vitamin D ) in hexane was measured at 280nm using 10mm cuvette.A 7.35 ppm calciferol solution gave a %T of 22.35%.
Given: molecular mass of calciferol is 396
a) What is the concentration of calciferol in fish sample in mg/g?
b)The absorbance of calciferol in another sample at 280nm is 0.25 when using 20mm cuvette .What is the concentration of this sample solution in ppm.
chemistry
2012-11-17 7:43 am
回答 (2)
2012-11-17 9:58 am
✔ 最佳答案
a)Concentration of calciferol in fish sample
= 7.35 ppm
= (7.35 mg) / (1000000 mg)
= (7.35 mg) / (1000 g)
= 0.00735 mg/g
b)
A = εbc, where
A : absorbance
ε : molar absorbtivity
b : path length of the sample
c : concentration of the sample
When A = -log(22.35%), b = 10 mm and c = 7.35 ppm :
A = εbc
-log(22.35%) = ε(10)(7.35)
ε = [-log(22.35%) / 73.5 ...... [1]
When A = 0.25, b = 20 mm :
A = εbc
0.25 = ε(20)c
ε = 0.0125 / c ...... [2]
For dilute solutions of calciferol, ε is a constant. Hence, [1] = [2] :
[-log(22.35%) / 73.5 = 0.0125 / c
c = 0.0125 x 73.5 / (-log22.35%)
Concentration of this sample solution, c = 1.41 ppm
參考: 賣女孩的火柴
2012-11-18 3:47 am
There're 3 varieties of "ppm".
ppm w/w = mass of solute / mass of solution
ppm w/v = mass of solute / volume of solution (unit is neglected)
ppm v/v = volume of solute / volume of solution (can be applied to gaseous mixture as well)
ppm w/w = mass of solute / mass of solution
ppm w/v = mass of solute / volume of solution (unit is neglected)
ppm v/v = volume of solute / volume of solution (can be applied to gaseous mixture as well)
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