1.

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.

Answer»

Solution :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 20g of ethylene glycol and 80 g of water.
Molar mass of `C_(2)H_(6)O_(2)=12xx2+1xx6+16xx2`
`= 62 g mol^(-1)`
Moles of `C_(2)H_(6)O_(2)=(20g)/(62g mol^(-1))=0.322` mol
Moles of water `= (80g)/(18g 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 CALCULATED as :`1-0.068=0.932`


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