5. A stock solution of red dye (molar mass 496.42 g/mol) is made by dissolving 0

Question

5. A stock solution of red dye (molar mass 496.42 g/mol) is made by dissolving 0.1049 g of red dye into a 50-mL volumetric flask and diluting the flask to the mark with distilled water. Then 15-mL of the stock solution is transferred into a 100-mL volumetric flask that is filled to the mark with distilled water. What is the concentration (in Molarity) of the final solution of red dye? Calculate the percent relative inherent error associated with the final solution of red dye. Note that 15-mL was obtained by using both a 5-mL and 10-mL volumetric pipet. What range of concentrations should be reported for the final solution based on the inherent error of the equipment? a. b. c.

Explanation / Answer

The initial concentration = 0.1049 g / 50mL

The molecular weight of dye = 496.42g / mole

Moles of dye / 50mL = 0.1049 / 496.42 = 2.11 X 10^-4 moles /50mL

therefore moles in 1mL = 4.23 X 10^-6

Moles in 1L = 0.00423

Therefore molarity = 0.00423 M

The volume taken of 0.00423 M solution = 15mL

Diluted to final volume = 100mL

Final concentration of dye solution= 0.00423 X 15 / 100 = 0.000635 Molar

b) error not given

c) errot not given

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