Question

calculate the volume in milliliters of a 0.42 m barium chlorate solution that contains 25.0 g of barium chlorate ($\text{ba(clo}_3\text{)}_2$). round your answer to 2 significant digits. $square$ ml

Explanation:

Step1: Calculate molar mass of $\text{Ba(ClO}_3\text{)}_2$

Molar mass = $137.33 + 2\times(35.45 + 3\times16.00) = 304.23\ \text{g/mol}$

Step2: Find moles of solute

$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{25.0\ \text{g}}{304.23\ \text{g/mol}} \approx 0.08217\ \text{mol}$

Step3: Calculate volume in liters

$\text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.08217\ \text{mol}}{0.42\ \text{mol/L}} \approx 0.1956\ \text{L}$

Step4: Convert to milliliters

$\text{Volume (mL)} = 0.1956\ \text{L} \times 1000 = 195.6\ \text{mL}$

Step5: Round to 2 significant digits

$195.6\ \text{mL} \approx 2.0\times10^2\ \text{mL}$

Answer:

$2.0\times10^2$ mL