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27 May, 20:34

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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  1. 28 May, 00:06
    0
    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
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