The percentage composition by mass of a solution is 20% urea `(NH_(2) CONH_(2))`, 40% glucose `(C_(6) H_(12) O_(6))`, and 40% water `(H_(2) O)`. Calculate the mole fraction of each component of the solution.

Explanation

Molar mass of urea `(NH_(2) CONH_(2))(Mw_(2))`
`(NH_(2))_(2) CO`
`= 2 (14 + 2) + 12 + 16`
`= 32 + 12 + 16 = 60 g mol^(-1)`.
Molar mass of glucose `(C_(6) H_(12) O_(6))(Mw_(3))`
`= 12 xx 6 + 12 + 6 xx 16`
`= 72 + 2 + 96 = 180 g mol^(-1)`
Molar mass of `H_(2) O (Mw_(1)) = 18 g mol^(-1)`
`X_(urea) (X_(2)) = (n_(2))/(n_(1) + n_(2) + n_(3)) = (W_(2) // Mw_(2))/((W_(1))/(Mw_(1)) + (W_(2))/(Mw_(2)) + (W_(3))/(Mw_(3)))`
`= (20 // 60)/((40)/(18) + (20)/(60) + (40)/(180))`
`(0.33)/(2.2 + 0.33 + 0.2)`
`= (0.33)/(2.75) = 0.12`
`X_(urea) (X_(3)) = (n_(3))/(n_(1) + n_(2) + n_(3)) = (W_(3) // Mw_(3))/((W_(1))/(Mw_(1)) + (W_(2))/(Mw_(2)) + (W_(3))/(Mw_(3)))`
`(40//180)/(2.2 + 0.33 + 0.22)`
`(0.22)/(2.2 + 0.33 + 0.22)`
`= (0.22)/(2.75) = 0.08`
`XH_(2) O (X_(1)) = [1 - (X_(2) + X_(2))] = [1 - (0.12 + 0.08)] = 0.8`
Hence, `X_(1) = 0.8, X_(2) = 0.12, X_(3) = 0.08`