Answer
a) $ 0.927 \space of \space water (amount)$
b) $ 3.37 \space M \space CH_3OH$
c) $4.34 \space m \space CH_3OH$
Work Step by Step
Mass of methanol $11.3 \space mL \times \frac{0.793 \space g }{1 \space mL} = 8.96 \space g$
Amount of methanol: $8.96 \space g\times \frac{1 \space mole}{32.04 \space g} = 0.280 \space mole$
Mass of solution = $75.0 \space mL \times \frac{0.980 \space g}{1 \space mL} = 73.5 \space g$
Mass of water = 73.5 g - 8.96 g = 64.5 g of water.
Amount of water = $64.5 \space g\times \frac{1 \space mole}{18.02 \space g} = 3.58 \space mole$
a) $\frac{3.58 \space mole}{0.280 \space mole + 3.58 \space mole} = 0.927$
b) $\frac{0.280 \space mole \space CH_3OH}{0.075 \space L \space solution} = 3.37 \space M$
c) $\frac{0.280 \space mole \space CH_3OH}{0.0645 \space kg \space water} = 4.34 \space m$