Question

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.

I got .582 M for K, 1.002 M for Cl, and .21 M for Ca. For some reason, masteringchem is saying that the first two answers are slightly off, but when i redid the problem, i didnt round anything until the end. Help!!!

Answer 1

\[Number of Moles=Mass \div Molar Mass\]
\[N of KCl=3.40\div39+35.5\]
\[N of KCl=0.045 Moles\]
\[N of moles=Concentration(mol/dm^3)*Volume(dm^3)\]
\[N of CaCl2=75*0.210\div1000\]
\[N of CaCl2=0.01575 Moles\]
\[CaCl2-->Ca+2+2Cl-
Calculate the concentration of each ion
K+=0.045
Cl-=0.045
Ca+2=0.01575
Cl-=0.01575*2
Total number moles of Cl-=(0.01575*2)+0.045
Number of Moles/Volume=Concenration
0.045/0.075=moles of K+
(0.01575*2)+0.045/0.075=moles of Cl-
0.01575/0.075= moles of Ca+2

Total Number of Moles present in Solution=0.06075

Answer 2

@kostek forget about the last part the "0.06075