Calculate the mole fraction of ethylene glycol `(C_(2)H_(6)O_(2))` in a solution containing `20%` of `C_(2)H_(6)O_(2)` by mass.

Explanation

i) Mole fraction is the ratio of number of moles of one component to the total number of moles of all the components in the solution.
Mole fraction of a component `= (" Number of moles of the component ")/(" Total number of moles of all the components in the solution ")`
ii) Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water.
Molar mass of `C_(2) H_(6) O_(2) = 12 xx 2 + 1 xx 6 + 16 xx 2 = 62 g mol^(-1)`.
Moles of `C_(2) H_(6) O_(2) = (20 g)/(62 g mol^(-1)) = 0.322 mol.`
Moles of water = ` (80 g)/(18 g mol^(-1)) = 4.444` mol
`X_("glycol") = ( " moles of " C_(2) H_(6) O_2)/( " moles of " C_(2) H_(6) O_(2) + " moles of " H_(2)O)`
`= (0.322 mol )/(0.322 mol + 4.444 mol) = 0.068`
Similarly, `X_( "water" ) = (4.444 mol)/(0.322 mol + 4.444 mol) = 0.932`
Mole fraction of water can also be calcu-lated as : `1 - 0.068 = 0.932.`