The mole fraction of `CH_(3)OH` in an aqueous solution is 0.02 and density is `94 "g cm"^(-3)`, Determine the molality of the solution.

Explanation

Step I. Calculation of molality of the solution
Molality of `CH_(3)OH` solution is number of moles of `CH_(3)OH` dissolved in 1000 g of water (kg of water)
No. of moles of `H_(2)O(n_(1))=((1000g))/(("18 g mol"^(-1)))=55.55 "mol"`
Mole fraction of `CH_(3)OH(x_(CH_(3)OH))=0.02`
No. of moles `CH_(3)OH(n_(2))` can be calculated as :
`(x_(CH_(3)OH))=(n_(2))/(n_(1)+n_(2))=(n_(2))/(55.55+n_(2))=0.02`
`n_(2)-0.02n_(2)=55.55xx0.02=1.111`
`0.98n_(1)=1.111or n_(2)=(1.111)/(0.98)=1.13` mol
`:.` Molality of `CH_(3)OH` solution `= 1.13` molal.