1.032 g of drink mix containing FD&C; Red #40 (496.42 g/mol) was dissolved in wa
ID: 883204 • Letter: 1
Question
1.032 g of drink mix containing FD&C; Red #40 (496.42 g/mol) was dissolved in water to form 100.0 ml solution. 2.00 ml of the stock solution was diluted to 100.0 ml with distilled water. The absorbance for the diluted drink mix solution is 0.427. Using the calibration curve below, calculate (a) the concentration of FD&C; Red #40 in the diluted unknown. (b) Calculate the %(m/m) FD&C; Red #40 in the drink mix. (a) [FD&C; Red #40]diluted = _________ mol/L (b) %FD&C; Red #40 = %(m/m)Explanation / Answer
a) FD&C red #40] dil=0.000017mol/L (located at x-axis for absorbance 0.427(y-ordinate) from the curve)
b) molar mass of fd&c red 40=496.42g/mol
concentration of drink mix=1.032g/100ml
After further dilution, C(stock solution)*V(stock)=C(diluted)* V(dil)
C(diluted)=C(stock solution)*V(stock)/V(dil)=1.032g/100ml* 2.0ml/100ml=0.00020 g/ml
concentration of FD&C red #40=0.000017mol/L*496.42g/mol=0.0084g/L=0.0084g/1000ml=0.0000084g/ml
%FD&C red #40]=concentration of FD&C red #40/actual concentration of drink mix *100
=0.0000084g/ml/0.00020 g/ml *100=4.2%(m/m)
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