Question

you have 150.0 ml of a 0.507 m solution of ce(no₃)₄. what mass (in grams) of ce(no₃)₄ would be required to make the solution? answer: g

Explanation:

Step1: Convert volume to liters

$V = 150.0\ \text{mL} = \frac{150.0}{1000} = 0.1500\ \text{L}$

Step2: Calculate moles of solute

Moles = Molarity × Volume
$n = 0.507\ \text{mol/L} \times 0.1500\ \text{L} = 0.07605\ \text{mol}$

Step3: Find molar mass of $\text{Ce(NO}_3\text{)}_4$

Molar mass of Ce = 140.12 g/mol, N = 14.01 g/mol, O = 16.00 g/mol
$M = 140.12 + 4\times(14.01 + 3\times16.00) = 140.12 + 4\times62.01 = 140.12 + 248.04 = 388.16\ \text{g/mol}$

Step4: Calculate mass of solute

Mass = Moles × Molar mass
$m = 0.07605\ \text{mol} \times 388.16\ \text{g/mol}$

Answer:

29.52 g