Question

How do you calculate the mass of lead (II) nitrate, `Pb(NO_3)_2`, necessary to make 50.0 mL of a 0.100 M solution?

Answer 1

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 L`

Now, `0.1M = n/(0.05L)` where `n` are the no. of moles of `Pb(NO_3)_2`

`n = 0.005` moles

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

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

Given mass `= 331 (gm)/(mol) xx 0.005 mol`

So mass of `Pb(NO_3)_2 = 1.655 gm`