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

Answer:

0.22

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{46.85\ g}{299.4\ g/mol}$

$Moles\ of\ C_{18}H_{21}NO_3= 0.1565\ mol$

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{125.5\ g}{46.07\ g/mol}$

$Moles\ of\ C_{2}H_{5}OH= 0.5535\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ codeine=\frac {n_{codeine}}{n_{codeine}+n_{ethanol}}$

$Mole\ fraction\ of\ codeine=\frac{0.1565}{0.1565+0.5535}=0.22$