Question

The solubility of BaCl2 in water is 43.6 g per 100 mL of water at 50 oC. Use this information to answer question belowCalculate the minimum amount of BaCl2 that must be added to 230mL of water at 50oC to make a saturated solution.

Answer 1

Explanation:

We are given: mass of BaCl2 = 43.6g

: volume of BaCl2 = 100mL

: volume of BaCl2 = 230mL

We know: molar mass of BaCl2 =208.23g/mol

We determine the number of moles of BaCl2:

$\begin{gathered} n\text{ = }\frac{m}{M} \\ \\ \text{ = }\frac{43.6}{208.23} \\ \\ \text{ = 0.21 mol} \end{gathered}$

We then find the concentration of BaCl2:

$\begin{gathered} n\text{ = CV} \\ \\ \therefore\text{ C = }\frac{n}{V} \\ \\ \text{ = }\frac{0.21}{0.1} \\ \\ \text{ = 2.09 M} \end{gathered}$

We then find the number of moles for a 230mL solution:

$\begin{gathered} n\text{ = CV} \\ \\ \text{ = 2.09}\times0.23 \\ \\ \text{ = 0.48 mol} \end{gathered}$

We then find the required mass:

$\begin{gathered} m\text{ = nM} \\ \\ \text{ = 0.48}\times208.23 \\ \\ \text{ = 100.28g} \end{gathered}$

Answer:

100.28g must be added per 230mL of water.