How do you calculate the mass of lead (II) nitrate, Pb(NO_3)_2Pb(NO3)2, necessary to make 50.0 mL of a 0.100 M solution?

1 Answer
Apr 27, 2017

1.655 gm

Explanation:

First, we'll look at formula for Molarity.

Molarity = No. of moles of solute / Vol. of solution (in L)

Before putting values, it helps to know that 50 mL = 0.05 L50mL=0.05L

Now, 0.1M = n/(0.05L)0.1M=n0.05L where nn are the no. of moles of Pb(NO_3)_2Pb(NO3)2

n = 0.005n=0.005 moles

Now we also know that n =n= given mass / molar mass

Molar mass of Pb(NO_3)_2 = 207 + 2(14 + 3(16)) = 331 (gm)/(mol)Pb(NO3)2=207+2(14+3(16))=331gmmol

Given mass = 331 (gm)/(mol) xx 0.005 mol=331gmmol×0.005mol

So mass of Pb(NO_3)_2 = 1.655 gmPb(NO3)2=1.655gm