Answer
a. $0.150 \space M \space N{O_3}^- $
b. $0.300 \space M \space N{O_3}^-$
c. $0.450 \space M \space N{O_3}^-$
Work Step by Step
1. Determine the amount of $N{O_3}^-$ ions in each compound:
a. 1
b. 2
c. 3
2. Use them as conversion factors:
a. $$0.150 \space M \space KNO_3 \times \frac{1 \space mol \space N{O_3}^-}{1 \space mol \space KNO_3} = 0.150 \space M \space N{O_3}^- $$
b. $$0.150 \space M \space Ca(NO_3)_2 \times \frac{2 \space mol \space N{O_3}^-}{1 \space mol \space Ca(NO_3)_2} = 0.300 \space M \space N{O_3}^- $$
c. $$0.150 \space M \space Al(NO_3)_3 \times \frac{3 \space mol \space N{O_3}^-}{1 \space mol \space Al(NO_3)_3} = 0.450 \space M \space N{O_3}^- $$
