Here is the following question:
Pb(NO3)2 (aq) + 2 KI (aq) → 2 KNO3 (aq) + PbI2(s)
If 20.50 (±0.05) mL of a 0.00105 M (±0.00002 M) KI stock solution that was diluted by 100, was required to create the yellow precipitate, what is the mass of the lead(II) nitrate in the sample?
I have gone up to the step on finding the mass of the lead(II) nitrate, however what I do not understand is that the moles that I have calculated for lead(II) nitrate has an uncertainty, but the molar mass that I have to look up does not have any uncertainty. Could someone explain what I should do in this case. I have attached 2 screenshots of what I have done with Part2.png following after Part1.png. Thanks!