Suppose `5 g` of acetic acid are dissolved in one litre of ethanol. Assume no reaction in between them. Calculate molality of resulting solution if density of ethanol is `0.789//mL`.

Explanation

Wt. of `CH_(3)COOH` dissolved `= 5 g`
Eq. of `CH_(3)COOH` dissolved `=5//60`
Volume fo ethanol `= 1 "litre"=1000 mL`
`:.` Weight of ethanol `=(1000xx0.789)= 789 g`
`:. "Molality of solution"=("Moles of solute")/("wt.of solvent in kg")`
`={(5)/((60xx789)/(1000))}=0.1056`