question 8 week 5

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Husnia Safi - 1K
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Joined: Fri Sep 29, 2017 7:04 am

question 8 week 5

Postby Husnia Safi - 1K » Wed May 09, 2018 11:28 am

Anhydrous calcium chloride, CaCl2, is a salt which is sued increasing water hardness in swimming pools. A solution of CaCl2 is prepared by dissolving 4.00g of salt to 1.00L water.
a) what is the concentration of chloride ions for the above solution?
b) The above solution diluted with 1.00L of 0.400M potassium chloride, KCl, solution. What is the concentration of chloride ions after dilution?

Im not even sure where to start

Celeste Martinez 1K
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Joined: Fri Apr 06, 2018 11:04 am
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Re: question 8 week 5

Postby Celeste Martinez 1K » Wed May 09, 2018 9:52 pm

a) Divide the 4.00 g of salt by the molar mass of CaCl2 (110.98 g/mol) = 0.036 mol of CaCl2.
Since we only want the concentration of chloride ions, we multiply 0.036 mol of CaCl2 by the 2 moles of chloride ions which gives us 0.072 mol Cl/1 L. Concentration of chloride ions is 0.072 M.
b) The total volume is 2 L since CaCl2 got diluted with 1 L of 0.400 M KCl
0.400 mol/1 L= 0.400 M
(0.400 + 0.072)/ 2 L = 0.236 M Cl

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