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Indicate the concentration of each ion present in the solution formed by mixing: 3.40g KCl in 75.0 mL of 0.210 M CaCl2 solution. Assume that the volumes are additive.

Verified Answer :

To solve this task, we have to use this molar formula : Number of Moles=Mass/Molar Mass.
According to this, we can calculate each : 
\(KCl=3.40/39+35.5\) which is 0,045 moles.
Then use this formula to go on :
№of moles=Concentration(mol/dm3)∗Volume(dm3), and  then \(CaCl2=75*0.210/1000\), which is 0.01575Moles.
Then you can calculate concentration :
\(K+=0.045 Cl-=0.045\)
\(Ca+2=0.01575 Cl-=0.01575*2\)
Total number : \(Cl-=(0.01575*2)+0.045\)
Here are moles of K+ 0.045/0.075
Of Cl- 
(0.01575*2)+0.045/0.075
Ca2+ 
0.01575/0.075
And the needed answer is definitely : 
=0.06075