Answer
125 mL of $0.150$ M $Li_2S$
Work Step by Step
1. Use the molarities and other information as conversion factors:
$$ 125 \space mL \space Co(NO_3)_2 \times \frac{1 \space L}{1000 \space mL} \times \frac{ 0.150 \space mol \space Co(NO_3)_2 }{1 \space L \space Co(NO_3)_2 } \times \frac{ 1 \space mol \space Li_2S }{ 1 \space mol \space Co(NO_3)_2 } \times \frac{1 \space L}{ 0.150 \space mol \space Li_2S } \times \frac{1000 \space mL}{1 \space L} = 125 \space mL $$
