Question

We need to prepare an aqueous solution containing 250.0 ppb of iodide. How many grams of potassium iodide (KI) would be required to prepare 1.000 L of this solution?

a) 250.0 μg b) 0.3270 mg c) 250.0 ng d) 327.0 ng e) other

Answer 1

Solution :-

1 ppb is 1 microgram per liter

so 250.0 ppb means 250 microgram

we need to make 1 L solution

so mass of the iodine needed = 250 microgram

so now

lets find the molar mass if KI

molar mass of KI = 166 g per mol

which contains 127 g Iodine

250 microgram * 1 g / 1*10^6 ug = 0.00025 g iodine

0.00025 g I * 167 g KI/ 127 g I = 0.000328 g KI

lets convert g to mg

0.000328 g * 1000 mg / 1 g = 0.328 mg

the closest answer is option B that 0.327 mg

So the answer is option b.