Mole fraction of water `chi_(H_(2)O) = 0.88`
Mole fraction of ethanol, `chi_(C_(2)H_(5)OH) = 1-0.88 = 0.12`
`chi_(C_(2)H_(5)OH) = (n_(2))/(n_(1) + n_(2)) " " ...(1)`
`n_(2) = "number of moles of ethanol"`
`n_(1) = "number of moles of water"`
Molality of ethanol means the number of moles of ethanol present in 1000 g of water.
`n_(1) = (1000)/(18) = 55.5 "moles"`
Substituting the value of `n_(1)` in equation (1)
`(n_(2))/(55.5+n_(2)) = 0.12`
`n_(2) = 7.57 "moles"`
Molality of ethanol `(C_(2)H_(5)OH) = 7.57 m`
Alternatively,
Mole fraction of water = 0.88
Mole fraction of ethanol = 1-0.88 = 0.12
Therefore, 0.12 moles of ethanol are present in 0.88 moles of water.
Mass of water `=0.88 xx 18 = 15.87`g of water
Molality = Number of moles of solute (ethanol) present in 1000 g of solvent (water)
`=12 xx 1000//15.84 = 7.57m`
Molality of ethanol `(C_(2)H_(5)OH) = 7.57m`
