Question

1. Serial dilution Questions: a) (5 pts) Suppose you begin with stock solution of dye that has a concentration of 350.0 mg/L and do the following set of serial dilutions on it: 5.00 mL of stock diluted to a total volume (TV) of 500.00 mL (Dilution 1); 2.00 mL of Dilution 1 is then diluted to a TV of 250.00 mL (Dilution 2); finally, 1.00 mL of Dilution 2 is then diluted to a total volume of 100.00 mL. Calculate the concentration of dye in this final solution in mg/L and mg/mL.

Answer 1

An concentration of Stock solution , m = 350mg 12 Dilution I concentration of solution after dilution 1= my volume of solution (diluted) = 4 Total volume of solution LV,J = 500 ml volunce of Stock solution taken (v) = 5.00mL using .concentration equation - (mv)etote = (M, VJditutions 350 mg/L x 5 mL = mi x 500 ml mi = 3.5mg / eBL / M = 3. Dilution 2 les conc. of Soph after délution 2 = m₂ Total volume of diluded Sol? = ₂ = 250 mL volume of dilution I sol" taken (v.) = 2ml

usuf (mio - (M₂ V2) ditutions 3.5 mg/1 x 2 = M₂ x 250 M₂ = 0.028 mg/l Final dilution, let couc. = M3 (final solution) Total volume of final solution = U₂ = 100ml volunce of Dilution 2 son taken = V₂ - 1.00mL usaq (Ma Va) ditution 2 My Vy final direction 0.028x8 = M3 x 100 conc. of dye M3 = 0.00028 mg/L/ after final ditution My0 00028 mg fent M = 0.000 28 mg / x 1000 = 0.28 mg/mu conc. = m.g of solette x1000 mgrouw of solute vol. in L Vol. in ML
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