Question

Exp 8 Pre-Lab Questions Name: Section: 140 6. A chemist takes a 100 mg/L vitamin C sample and exposes it to the open air for two weeks. The concentration of vitamin C in the exposed sample was reduced to 75 mg/L. Calculate the % decrease in vitamin C concentration in the original sample. Show your calculation 75/100 = 1 meaning that there was a 25% decrease In organd sample concetton When doing dilutions whilst working in concentration of mg/L, remember CIVI-C2V2 where C is concentration in mg/L and V is volume in mL. CI will be the concentration of the solution you are diluting (our master stock of vitamin C which is 40mg/L), C2 will be the desired concentration of the dilution, V1 the amount diluted (pipette size generally) and V2 the amount of the diluted solution desired (flask size). C1 and C2 are dictated by what solutions you are g so you will choose either V1 or V2 and the expression CIVI - C2V2 will solve to give you the other. Complete the table below PRIOR to coming to lab. Name of vitamin C std Concentration in mg/L Volume of Stock Vitamin C solution needed (mL) Final volume of standard solution (mL)

Answer 1

#6. Initially concentration = 100mg/L

Concentration after exposure = 75mg/L

Decrease in concentration = 100 - 75 = 25mg/L

Percent decrease in concentration = (decrease in concentration/initial concentration)×100 = (25/100)×100 = 25% decrease in concentration.

Yours answer is correct.