We have 0.20 mol/L of solution and we required 750.0 mL
Concentration= 0.20 mol/L
Volume = 750.0 mL = 0.7500 L
If we multiply,
[tex]0.20\text{ }\frac{mol}{L}x0.7500L\text{ = 0.15 moles}[/tex]After this, we need to find the mass of these 0.15 moles, so we need the molar mass of lead (II) nitrate:
The molar mass = 331 g/mol
Now,
[tex]0.15\text{ moles x 331 }\frac{g}{\text{mol}}=49.6\text{ g = 50 g (approx.)}[/tex]Answer: 50 g
