## Chemistry: Molecular Approach (4th Edition)

a. $0.200 \space M \space Cl^-$ b. $0.300 \space M \space Cl^-$ c. $0.300 \space M \space Cl^-$
1. Determine the amount of $Cl^-$ ions in each compound: a. 1 b. 2 c. 3 2. Use them as conversion factors: a. $$0.200 \space M \space NaCl \times \frac{1 \space mol \space Cl^-}{1 \space mol \space NaCl} = 0.200 \space M \space Cl^-$$ b. $$0.150 \space M \space SrCl_2 \times \frac{2 \space mol \space Cl^-}{1 \space mol \space SrCl_2} = 0.300 \space M \space Cl^-$$ c. $$0.100 \space M \space AlCl_3 \times \frac{3 \space mol \space Cl^-}{1 \space mol \space AlCl_3} = 0.300 \space M \space Cl^-$$