The mole fraction of `CH_(3)OH` in an aqueus solution is 0.02 and density is `0.994 g cm^(-3)`. Determine the molality and molarty.

Explanation

Calculation of molality of the solution.
Molality of `CH_(3)OH` solution is number os moles of `CH_(3)OH` dissoved in 1000 g of water(kg of water)
No. of moles of `H_(2)O(n_(1))=((1000g))/((18gmol^(-1))=55.55 mol`
Mole fraction of `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)/(0.98)=1.13 mol`
Molality of `CH_(3)OH` solution (m)1.13 molal
Calculation of molarity of the slution.
Molarity of soultion (M)`=("No.of gram moles of" CH_(3)OH)/("Volume of solution is in litre")`
In solution, mass of `H_(2)O=1000g, "mass of" CH_(3)OH=(1.13mol)xx(32 g mol^(-1)=36.16 g)`
Total mass of solution = 0.994 g `cm^(-3)`
Volume of solution`=("Mass of solution")/("Density") =((1036.16g))/((0.994 g cm^(-3))=1042.4cm^(3)`
Molarity of solution (M)`=((1.13mol))/(((1042.4)/(1000)dm^(3)))=((1.13xx1000mol))/((1042.4dm^(3)))`
`=1.08mol (dm^(3))^(-1)=1.08 M`