# 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?

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 $50 m L = 0.05 L$

Now, $0.1 M = \frac{n}{0.05 L}$ where $n$ are the no. of moles of $P b {\left(N {O}_{3}\right)}_{2}$

$n = 0.005$ moles

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

Molar mass of $P b {\left(N {O}_{3}\right)}_{2} = 207 + 2 \left(14 + 3 \left(16\right)\right) = 331 \frac{g m}{m o l}$

Given mass $= 331 \frac{g m}{m o l} \times 0.005 m o l$

So mass of $P b {\left(N {O}_{3}\right)}_{2} = 1.655 g m$