Question

1. how many grams of copper(ii) sulfate pentahydrate ($\text{cuso}_4 cdot 5\text{h}_2\text{o}$) are needed to prepare 100.00 milliliters of a 0.12 $m$ copper(ii) sulfate solution?

Explanation:

Step1: Convert volume to liters

$V = 100.00\ \text{mL} = \frac{100.00}{1000} = 0.10000\ \text{L}$

Step2: Calculate moles of solute

$n = M \times V = 0.12\ \text{mol/L} \times 0.10000\ \text{L} = 0.012\ \text{mol}$

Step3: Find molar mass of $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$

Molar mass = $63.55 + 32.07 + 4\times16.00 + 5\times(2\times1.01 + 16.00)$
$= 63.55 + 32.07 + 64.00 + 5\times18.02 = 249.72\ \text{g/mol}$

Step4: Calculate mass of solute

$m = n \times \text{molar mass} = 0.012\ \text{mol} \times 249.72\ \text{g/mol}$

Answer:

$2.99664\ \text{g}$ (or $\approx 3.0\ \text{g}$ when rounded to two significant figures)