2a) Put two drops of hydroxylamine solution in a 0.50 mL well. Then add 8 drops

Question

2a) Put two drops of hydroxylamine solution in a 0.50 mL well. Then add 8 drops of 1.00 g L^-1 bipyridine and two drops of 5 mg L^-1 iron. How many moles of bipyridine and how many moles of iron are in the well? There are about 20 drops to 1 mL.

2b) If you want to repeat the mixture from part a on a larger scale, describe how many milliliters of hydroxylamine solution, how many milliliters of bipyridine solution, and how many milliliters of iron solution you need to make 2.50 mL of total solution volume?

PLEASE SHOW ALL WORK

Explanation / Answer

a)

vol. of iron = 5/20 = 0.25 mL

vol. of bipyridine = 8/20 = 0.4 mL

so,

wt of iron = 5*10^-3 * 0.25*10^-3 = 0.00000125 g

wt of bipyridine = 1* 0.4*10^-3 = 0.0004 g

so,

no. of moles of iron = 0.00000125/55.845 = 2.23833826e-8 moles

moles of bipyridine = 0.0004/156.18 = 0.00000256114 moles

b)

total volume original in mixture = 2/20+8/20+5/20 = 0.75 mL

milliliters of hydroxylamine solution = (2/20)*2.5/0.75 = 0.333333 mL

milliliters of bipyridine solution = (8/20)*2.5/0.75 = 1.333333 mL

milliliters of iron solution = (5/20)*2.5/0.75 = 0.8333 mL

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