Question

How many grams of Ca(CH3COO)2 (calcium acetate) are in 175.8 mL of a 0.207 M Ca(acetate)2 solution
A 2.3 x 10
B 3.64 x 102
C 3.364
5.75
E None of these الاجابه بالعربي

Answer 1

The molecular weight of Ca(CH3COO)2 is 158.17 g/mol.
First, we need to calculate the number of moles of Ca(CH3COO)2 in 175.8 mL of a 0.207 M solution:
0.207 moles/L * 0.1758 L = 0.03612 moles
Next, we can calculate the grams of Ca(CH3COO)2:
0.03612 moles * 158.17 g/mol = 5.71 grams
Therefore, the answer is approximately 5.71 grams. So, the closest answer is 5.75.