I. A solution is prepared by dissolving 0.1896 grams of FD&C; Red #40 in water t
ID: 568914 • Letter: I
Question
I. A solution is prepared by dissolving 0.1896 grams of FD&C; Red #40 in water to CisHIAN S Os determine the molarity of this stock solution. e 2.500 Liters of solution. Given that the molecular formula of Red 40 is 4S0.4 cters o.ccomol |2.5L 450.4 = o.ooo 1084 M 0.000 Ib84M 2, 25.00 m,L of the dark red stock solution described question #1 is transferred to a volumetric flask and diluted with water to 100.00 mL a. how many moles of Red 40 are in the 25.00 mL sample that was transferred? nolanty >Molers (0.000lb84A)(0.02 bl) Molmolanty xaters o.ooooo42 to 4.210 XO b. After the dilution, what is the molarity of the Red 40 in the 100.00 mL of solution? moles (o.o0oM(25mL) Ma C.ODoo y 21 M2 Cl0om) 1.21Dx1O-5 note: the preceding problem (#2) is a dilution calculation. A convenient formula for calculating concentrations after dilution is Miv, = My, where Mi and VI are the molarity and volume of the concentrated solution, respectively, and M2 and V2 are the molarity and volume of the dilute solution. 3.a. The solution described in problem #2 is placed in the spectrophotometer. Based on the calibration curve shown on the next page, what would you predict the absorbance to be? Note: like the example given in the discussion, the absorbance should be calculated, not just estimated from the graph. Notice that the slope and intercept of the calibration curve are provided in the figure on the next page. Albsorbance len ADSz(2.10330")(4.21 D X 10-5) Absorbance- = O.DB55Explanation / Answer
1.
we know molarity = (weight / molecular weight) * (1 / volume of the solution (lit))
molarity of this stock solution = (0.1896 / 496.92) * (1 / 2.5)
= 1.526 * 10-4 M
2.
a)
we are transferring 25 ml of 1.526 * 10-4 M of solution to another beaker.
therefore number of moles present in 25ml solution = 0.025 lit * 1.526 * 10-4M
= 3.81 * 10-6moles
b)
after dilution to 100 ml, then the
molarity of the solution = number of moles present in 25ml of solution / 0.1 lit = 3.81 * 10-6 moles / 0.1
= 3.81 * 10-5M