Question

An aqueous solution of ethanol, CH3CH2OH, has a concentration of 5.82 mol/L and has a density of 0.957 g/mL. What are the mass percent and mole fraction of CH3CH2OH in this solution

Answer 1

1. Mass percent of ethanol = 28.02 %.

2. Mole fraction of ethanol = 0.13.

Step-by-step explanation:

1. To find the mass percent we need to use the following equation:

`\%_(m/m) = (m_(e))/(m_(s)) * 100`

Where:

`m_(e)` is the mass of ethanol

`m_(s)` is the mass of the solution

We need to calculate the mass of ethanol and the mass of the solution:

`d = (m_(s))/(V)`

Where:

d: is the density of the solution = 0.957 g/mL

V: is the volume = 1 L

`m_(s) = d*V = 0.957 (g)/(mL)*(1000 mL)/(1 L)*1 L = 957 g`

Now, from the concentration we can find the mass of ethanol:

`C = (n_(e))/(V) = (m_(e))/(M_(e)*V)`

Where:

`M_(e)`: is the molar mass of ethanol = 46.07 g/mol

`n_(e)`: is the number of moles of ethanol = m/M

`m_(e) = C*M*V = 5.82 (mol)/(L)*46.07 (g)/(mol)*1 L = 268.13 g`

Finally, the mass percent of ethanol is:

`\%_(m/m) = (268.13 g)/(957 g) * 100 = 28.02 \%`

2. The mole fraction of ethanol is given by:

`\chi_(e) = (n_(e))/(n_(s))`

The number of moles of ethanol is:

`n_(e) = (m_(e))/(M_(e)) = (268.13 g)/(46.07 g/mol) = 5.82 moles`

And the moles of the solution is:

`n_(s) = n_(e) + n_(w)`

Where w is for water

`n_(s) = n_(e) + (m_(w))/(M_(w))`

`n_(s) = n_(e) + (m_(s) - m_(e))/(M_(w))`

`n_(s) = 5.82 moles + (957 g - 268.13 g)/(18 g/mol) = 44.09 moles`

Hence, the mole fraction of ethanol is:

`\chi_(e) =(5.82 moles)/(44.09 moles)=0.13`

I hope it helps you!