Question

Question 4

If you have a solution that is made by dissolving 345 grams of CaCO3 (molar mass 100.09 g/mol) to
make a solution with a final volume of 2.25L, what is the molarity of the calcium carbonate solution?
0.129 M
1.53 M
153 M
3.45 M

Answer 1

Answer: 1.53 M

Step-by-step explanation:

1) solve for moles

345 g CaCO3 x ( 1mol/100.09 g) =3.446897792 moles

2) plug in formula M=mole/liters

3.446897792 / 2.25 = 1.53 M