Q98P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate each of the following quantities: (a) Grams of solute needed to make 475 mL of 5.62×10^-2 M potassium sulfate (b) Molarity of a solution that contains 7.25 mg of calcium chloride in each milliliter (c) Number of Mg²⁺ ions in each milliliter of 0.184 M magnesium bromide.

Step-by-Step Solution

Answer

(a) 4.65 g K₂SO₄ was needed to make the solution of 475 mL of 5.62×10^-2 M potassium sulfate

(b) Molarity is found as $6.53\times {10}^{-2} M CaC{l}_2$

(c) The number of magnesium ions in each ml of 0.184 M Magnesium bromide is $1.11\times {10}^{20} M{g}^{2+}ions.$

a. Step 1: Finding the amount need to make 475 mL of 5.62×10 -2 M potassium sulfate

The solution volume and molar mass of potassium sulfate can be multiplied as,

$\begin{array}{l}Mass of K_{2}S{O}_4=475\times {10}^{-3}L so\ln \times \frac{5.62\times {10}^{-2}mol K_{2}S{O}_4}{L so\ln }\times \frac{174.26 g K_{2}S{O}_4}{1 mol K_{2}S{O}_4}\\ Mass of K_{2}S{O}_4=4.65 g\end{array}$

b. Step 2: Finding molarity

On multiplying the given calcium carbonate mass by molar mass reciprocal to find mole number present in solution.

$\begin{array}{l}Moles of CaC{l}_2=7.25\times {10}^{-3} g CaC{l}_2\times \frac{1 mol CaC{l}_2}{110.98 g CaC{l}_2}\\ =6.53\times {10}^{-5}mol CaC{l}_2\\ Molarity=\frac{6.53\times {10}^{-5}mol CaC{l}_2}{1\times {10}^{-3}Lso\ln }\\ Molarity=6.53\times {10}^{-2}M\end{array}$

c. Step 3: Finding the number of magnesium ions

The given volume of solution can be multiplied by molarity, the ratio between ions, and the Avogadro number to find the number of magnesium ions.

$\begin{array}{l}M{g}^{2+}ions=1\times {10}^{-3}L so\ln \times \frac{0.184 mol MgB{r}_2}{1 mol MgB{r}_2}\times \frac{1 mol M{g}^{2+}ions}{1 mol MgB{r}_2}\times \frac{6.022\times {10}^{23} M{g}^{2+}ions}{1 mol M{g}^{2+}ions}\\ M{g}^{2+}ions=1.11\times {10}^{20}M{g}^{2+}ions\end{array}$