Question

Calculate the mass of solid required to make 750.0mL of 0.20mol/L solution of lead ii nitrate

Answer 1

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,

`0.20\text{ }(mol)/(L)x0.7500L\text{ = 0.15 moles}`

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,

`0.15\text{ moles x 331 }\frac{g}{\text{mol}}=49.6\text{ g = 50 g (approx.)}`

Answer: 50 g