Question

If you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of codeine?

Answer 1

0.22

Step-by-step explanation:

Given, Mass of
`C_(18)H_(21)NO_3` = 46.85 g

Molar mass of
`C_(18)H_(21)NO_3` = 299.4 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (46.85\ g)/(299.4\ g/mol)`

`Moles\ of\ C_(18)H_(21)NO_3= 0.1565\ mol`

Given, Mass of
`C_(2)H_(5)OH` = 125.5 g

Molar mass of
`C_(2)H_(5)OH` = 46.07 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (125.5\ g)/(46.07\ g/mol)`

`Moles\ of\ C_(2)H_(5)OH= 0.5535\ mol`

So, according to definition of mole fraction:

`Mole\ fraction\ of\ codeine=\frac {n_(codeine)}{n_(codeine)+n_(ethanol)}`

`Mole\ fraction\ of\ codeine=(0.1565)/(0.1565+0.5535)=0.22`