Q3.100P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate each of the following quantities: (a) Molarity of a solution prepared by diluting 37.00 mL of 0.250 M potassium chloride to 150.00 mL (b) Molarity of a solution prepared by diluting 25.71 mL of 0.0706 M ammonium sulfate to 500.00 mL (c) Molarity of sodium ion in a solution made by mixing 3.58 mL of 0.348 M sodium chloride with 500. mL of 6.81×10^-2 M sodium sulfate (assume volumes are additive).

Step-by-Step Solution

Answer

1. The molarity of a solution prepared by diluting 37.00 mL of 0.250 M potassium chloride to 150.00 mL is 0.062 M HCl.
2. The molarity of a solution prepared by diluting 25.71 mL of 0.0706 M ammonium sulfate to 500.00 mL is $3.63\times {10}^{-3}M (N{H}_4{)}_{2}S{O}_4.$
3. The molarity of a sodium ion in a solution made by mixing 3.58 mL of 0.348 M sodium chloride with 500 mL of 6.81×10^-2 M sodium sulfate is 0.138 M Na⁺ ions.

Step 1: Finding the molarity of a solution prepared by diluting 37.00 mL of 0.250 M potassium chloride to 150.00 mL

a. On and solving,

$\begin{array}{rcl}{M}_{dil}& =& \frac{M_{conc}\times V_{conc}}{V_{dil}}\\ & =& \frac{0.250 M\times 37mL}{150 mL}\\ & =& 0.062 M KCl.\end{array}$

Step 2: Finding the molarity of a solution prepared by diluting 25.71 mL of 0.0706 M ammonium sulfate to 500.00 mL

b. On and solving,

$\begin{array}{rcl}{M}_{dil}& =& \frac{M_{conc}\times V_{conc}}{V_{dil}}\\ & =& \frac{0.0706 M\times 25.71mL}{500 mL}\\ & =& 3.63\times {10}^{-3} M (N{H}_4{)}_{2}S{O}_4.\end{array}$

Step 3: Finding the molarity of a sodium ion in a solution made by mixing 3.58 mL of 0.348 M sodium chloride with 500 mL of 6.81×10 -2 M sodium sulfate

c. Add the values to find the mole number of the Na⁺ ions.

$\begin{array}{rcl}Moles of N{a}^{+} ions (from NaCl)& =& 3.58\times {10}^{-3}L so\ln \times \frac{0.348 mol NaCl}{L so\ln }\times \frac{1 mol N{a}^{+}ions}{1 mol NaCl}\\ & =& 1.25\times {10}^{-3} mol N{a}^{+}ions.\\ & & \\ N{a}^{+} ions (from N{a}_{2}S{O}_4)& =& 500\times {10}^{-3}L so\ln \times \frac{6.81\times {10}^{-2}mol N{a}_{2}S{O}_4}{L so\ln }\times \frac{2 mol N{a}^{+}ions}{1 mol N{a}_{2}S{O}_4}\\ & =& 6.81\times {10}^{-2}mol N{a}^{+}ions.\\ & & \\ Moles of N{a}^{+}ions \left(total\right)& =& 1.25\times {10}^{-3}mol+6.81\times {10}^{-2}mol\\ & =& 0.06935 mol N{a}^{+}ions.\\ V_{total}& =& 3.58\times {10}^{-3}L+500\times {10}^{-3}L\\ & =& 0.50358 L.\end{array}$

Divide the number of moles calculated by the total volume of the solution to find the molarity of the Na⁺ ions.

$\begin{array}{rcl}Molarity& =& \frac{0.06935 mol N{a}^{+}ions}{0.50358 L so\ln }\\ & =& 0.138 M.\end{array}$