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How many grams of Lead(II) nitrate are present in 150.0 mL of an Lead(II) nitrate solution that is 0.07268 M?

How many grams of Lead(II) nitrate are present in 150.0 mL of an Lead(II) nitrate solution that is 0.07268 M?
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How many grams of Lead(II) nitrate are present in 150.0 mL of an Lead(II) nitrate solution that is 0.07268 M?

asked
User Ruchit Patel
by
8.1k points

1 Answer

4 votes
hey...
Use the molarity formula
M=moles/L and then convert to grams

0.07268*0.15=moles
0.010902 mol
Pb(NO3)2
1 mole=331.22g
0.010902 moles=

3.61 g


answered
User Marcumka
by
7.6k points
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