How do you calculate the mass of lead(II) nitrate, Pb{{(N{{O}_{3}})}_{2}} , necessary to make 50 mL of a 0.100M solution?

To calculate the mass of lead(II) nitrate \(\text{Pb(NO}_3\text{)}_2\) necessary to make 50 mL of a 0.100 M solution, follow these steps:

### Step 1: Use the molarity formula
Molarity (\(M\)) is defined as:
\[
M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
\]

We are given:
- Molarity \(M = 0.100 \, \text{mol/L}\)
- Volume of solution \(V = 50 \, \text{mL} = 0.050 \, \text{L}\)

First, calculate the moles of lead(II) nitrate required using the molarity formula:
\[
\text{Moles of Pb(NO}_3\text{)}_2 = M \times V
\]
\[
\text{Moles of Pb(NO}_3\text{)}_2 = 0.100 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{mol}
\]

### Step 2: Calculate the molar mass of lead(II) nitrate
The molar mass of \(\text{Pb(NO}_3\text{)}_2\) can be calculated by adding the atomic masses of each element:
- Pb: \(207.2 \, \text{g/mol}\)
- N: \(14.01 \, \text{g/mol}\)
- O: \(16.00 \, \text{g/mol}\)

So the molar mass is:
\[
\text{Molar mass of Pb(NO}_3\text{)}_2 = 207.2 + 2 \times (14.01 + 3 \times 16.00)
\]
\[
\text{Molar mass of Pb(NO}_3\text{)}_2 = 207.2 + 2 \times (14.01 + 48) = 207.2 + 2 \times 62.01 = 207.2 + 124.02 = 331.22 \, \text{g/mol}
\]

### Step 3: Calculate the mass of lead(II) nitrate required
Now that we have the number of moles (0.005 mol) and the molar mass (331.22 g/mol), we can calculate the mass:
\[
\text{Mass of Pb(NO}_3\text{)}_2 = \text{moles} \times \text{molar mass}
\]
\[
\text{Mass of Pb(NO}_3\text{)}_2 = 0.005 \, \text{mol} \times 331.22 \, \text{g/mol} = 1.6561 \, \text{g}
\]

### Final Answer:
The mass of lead(II) nitrate required to make 50 mL of a 0.100 M solution is approximately **1.66 g**.