Chapter 13 - Solutions and Their Physical Properties - Exercises - Molarity - Page 593: 19

1.85 M

1. Calculate the molar mass: $ CH_3OH $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 32.04 g/mol 2. Use the informations as conversion factors to find the molarity of this solution: $$\frac{ 6.00 g \space CH_3OH }{100 \space g \space solution} \times \frac{1 \space mol \space CH_3OH }{ 32.04 \space g \space CH_3OH } \times \frac{ 0.988 \space g \space solution}{1 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} = 1.85 \space M$$