Answer to Question #167322 in General Chemistry for Jose Ulises Ramirez

Question

A sample of an industrial waste water is analyzed and found to contain 44.9 ppb Mn²⁺. How many grams of manganese could be recovered from 1.17×10³Â kg of this waste water?

Answer 1

44.9ppb = 44.9 microgram per Kg

Hence, 1Kg of waste water contains 44.9 microgram of Mn2+

1.17x10^3Kg of Waste water contains (44.9x1.17x10^3)

= 52533microgram

Converting microgram to grams, since the question requested for grams

= 52533x10^-9 = 5.2533 x 10^-5g