Answer to Question #90577 in General Chemistry for acadia carlson

Question #90577
A solution was made by putting 350mL of a 2.5%m/v solution of potassium chloride with 250mL of a 0.250mol/L solution of magnesium chloride into a beaker. what is the concentration of chloride ion? (mol/L)
1
Expert's answer
2019-06-06T05:18:37-0400
  1. A mass of potassium chloride in 2.5% m/v solution found by using formula:


"2.5 \\%m\/v=\\frac{m KCl}{350mL}\\cdot100""mKCl = \\frac {2.5 \\%g\/mL \\cdot 350mL} {100}"

"mKCl=8.75 g"

2. Number of moles of the KCl is found by dividing the mass by molecular weight of this salt (74.55 g/mol):


"n=\\frac{m}{MW}=\\frac{8.75g}{74.55g\/mol}=0.117moles"

3. Number of moles of chloride ion in the solution of potassium chloride is:


"n=1\\cdot0.117=0.117moles"

4. To find a moles of MgCl2 one must multiply the molar concentration by it's volume in L:


"n=M \\cdot V""n= 0.250 \\frac{mol}{L} \\cdot 0.250 L = 0.0625moles"

5. number of moles of chloride ion in the solution of magneisum chloride is:


"n=2\\cdot 0.0625moles=0.125moles"

6. Total number of moles of chloride ion in the final solution:


"n= {0.117moles}+{0.125moles}=0.242 moles"

7. Total volume of final solution:


"V= \\frac{350mL+250mL}{1000}=0.6L"

8. Concentration of chloride ion:


"M=\\frac{n}{V}=\\frac{0.242moles}{0.6L}=0.4 \\frac{moles}{L}"


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