Question

Calculate the concentration of Ci' in a solution made by mixing 41.4 mL of 1.03 M NaCl with 66.2 mL of 0.818 M CaCl2 and diluting to 300.0 mL.

a) 02.02 M
b) 1.01M
c) 0.503 M
d) C: 0.361M
e) 0414M

Answer 1

The concentration of Cl- ions in the final solution is 0.503 M after calculating the total moles of Cl- provided by NaCl and CaCl2 and dividing by the total volume. Therefore, the correct option is c) 0.503 M.

Step-by-step explanation:

To calculate the concentration of Cl- ions in a solution made by mixing 41.4 mL of 1.03 M NaCl with 66.2 mL of 0.818 M CaCl2 and diluting to 300.0 mL, we must first calculate the number of moles of Cl- ions contributed by each salt, then add these together and divide by the total volume of the solution in liters.

From NaCl (1 Cl- per NaCl molecule):

• (41.4 mL)(1.03 mol/L) = 0.042642 mol of Cl-

From CaCl2 (2 Cl- per CaCl2 molecule):

• (66.2 mL)(0.818 mol/L) × 2 = 0.1082164 mol of Cl-

Total moles of Cl- = 0.042642 mol + 0.1082164 mol = 0.1508584 mol

Now we divide the total moles by the final volume in liters:

• (0.1508584 mol) / (0.300 L) = 0.503 M

Therefore, the concentration of Cl- in the solution is 0.503 M.