Question

An aqueous soulution contains 30% C3H7OH and 70% water by mass. what are the mole fractions of each substance in the solution?

Answer 1

Mole fraction of water = 0.88

Mole fraction of
`\chi_{C_(3)H_(7)OH}` = 0.12

Step-by-step explanation:

30%
`C_(3)H_(7)OH` means 30 g of
`C_(3)H_(7)OH` is present in 100 g of solution.

70%
`H_(2)O`(water) means 70 g of water is present in 100 g of solution

`n=(given\ mass)/(molar\ mass)`

Calculation of MOLES

Molar mass of
`H_(2)O` = 18 g/mol

`n_(water)=(70)/(18)`

`n_(water) = 3.89\ moles`

moles of water = 3.89 moles

Molar mass of
`C_(3)H_(7)OH` = 3(12) + 7(1) + 1(16) +1(1)

= 36 +7+16+1

molar mass of
`C_(3)H_(7)OH` = 60 g/mol

`n_{C_(3)H_(7)OH}=(30)/(60)`

`n_{C_(3)H_(7)OH} = 0.5\ moles`

moles of
`C_(3)H_(7)OH` = 0.5 moles

total moles of solution = 0.5 + 3.89 = 4.39 moles

Calculation of MOLE FRACTION

Formula for mole fraction is :

`\chi=(moles\ of\ solute)/(moles\ of\ solution)`

`\chi_(water)=(moles\ of\ water)/(Total\ moles)`

`\chi_(water)=(3.89)/(4.39)`

`\chi_(water)= 0.88`

Similarly,

`\chi_{C_(3)H_(7)OH} = (moles\ of\ C_(3)H_(7)OH)/(Total\ moles)`

`\chi_{C_(3)H_(7)OH} = (0.5)/(4.39)`

`\chi_{C_(3)H_(7)OH} = (0.5)/(4.39)`

`\chi_{C_(3)H_(7)OH} = 0.12`

Mole fraction of water = 0.88

Mole fraction of
`\chi_{C_(3)H_(7)OH}` = 0.12

Note :

`\chi_{C_(3)H_(7)OH} + \chi_{H_(2)O} = 1`