Calculate the molarity (M) and molality (m) of 16% aqueous methanol `(CH_3 OH)` solution by volume. (Given : Density of solution = 0.9 g m`L^(-1)`).

Explanation

16 mL of CH,OH is dissolved in 100 mL of solution.
Volume or mass of `H_2 O = 100 - 16 = 84 mL or 84 g`
`( :. d_(H_2 O) ~~ 1 g mL^(-1))`
Mass of solution `= V_(sol) xx d_(sol) = 100 xx 0.9 = 90 g `
Mass of solute `(w_2) = W_(s o l) - w_1 = 90 - 84 = 6 g`
Molar mass of `CH_3 OH = 32 g mol^(-1)`
` therefore M=(w_2 xx 1000 )/(M_2 xx V_("sol"))=(6 xx 1000 )/(32 xx 100 ) =1.875 M`
`m=(w_2xx 1000 )/(M_2 xx w_1 )=(6 xx 1000 )/(32 xx 84 )=2.232 m`