Question

Lead is a highly poisonous metal. The mass percentage of lead in the 250 g aqueous solution was found to be 1.2×10−6% by extracting the lead with an excess of iodide and weighing the resulting PbI2. How many micrograms (μg) of lead iodide were measured? Your answer should have two significant figures (round to the nearest tenth).

3 is not the answer

Answer 1

Given:-

mass percentage(%) of lead (Pb) = 1.2 10^-6

wt. of aqueous solution of lead = 250 g

wt of lead iodide (PbI₂) = ?

As we know that

molar mass of lead (Pb) = 207 g / mol

molar mass of iodine (I) = 127 g / mol

molar mass of lead iodide (PbI₂) = 207 + ( 2 127)

molar mass of lead iodide (PbI₂) = 207 + 254

molar mass of lead iodide (PbI₂) = 461 g / mol

therefore

mass of lead (Pb) present in 250 g aqueous solution = wt. of solution mass percentage(%) of lead (Pb) / 100

mass of lead (Pb) present in 250 g aqueous solution = 250 1.2 10^-6 / 100

mass of lead (Pb) present in 250 g aqueous solution = 300 10^-6 / 100

mass of lead (Pb) present in 250 g aqueous solution = 3.0 10^-6 g Pb

As we know that extraction of lead is carried out with help of excess iodine to form lead iodide (PbI₂).

So we know that

207 g lead (Pb) is present in = 461 g lead iodide (PbI₂)

then

1 g  lead (Pb) would be present in = 461/ 207 g lead iodide (PbI₂)

1 g  lead (Pb) would be present in = 2.2271 g lead iodide (PbI₂)

3.0 10^-6 g lead (Pb) would be present in = 2.2271 3.0 10^-6 g lead iodide (PbI₂)

3.0 10^-6 g lead (Pb) would be present in = 6.6812 10^-6 g lead iodide (PbI₂)

As we know that

1 micrograms (g) = 10^-6 g

therefore

3.0 10^-6 g lead (Pb) would be present in = 6.6812 micrograms (g)  lead iodide (PbI₂)

3.0 micrograms (g) lead (Pb) would be present in = 6.6812 micrograms (g)  lead iodide (PbI₂)

i.e the wt. of lead iodide (PbI₂) was measured = 6.6812 g

therefore answer is 6.6812 g lead iodide (PbI₂) was measured during extraction of lead (Pb) from 250 g aqueous solution.

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