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

Answer:

1. Mass percent of ethanol = 28.02 %.

2. Mole fraction of ethanol = 0.13.

   

Explanation:

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

$\%_{m/m} = \frac{m_{e}}{m_{s}} \times 100$

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:

$d = \frac{m_{s}}{V}$

Where:

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

V: is the volume = 1 L

$m_{s} = d*V = 0.957 \frac{g}{mL}*\frac{1000 mL}{1 L}*1 L = 957 g$

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

$C = \frac{n_{e}}{V} = \frac{m_{e}}{M_{e}*V}$                      

Where:

$M_{e}$: is the molar mass of ethanol = 46.07 g/mol

$n_{e}$: is the number of moles of ethanol = m/M

$m_{e} = C*M*V = 5.82 \frac{mol}{L}*46.07 \frac{g}{mol}*1 L = 268.13 g$

Finally, the mass percent of ethanol is:

$\%_{m/m} = \frac{268.13 g}{957 g} \times 100 = 28.02 \%$

2. The mole fraction of ethanol is given by:

$\chi_{e} = \frac{n_{e}}{n_{s}}$

The number of moles of ethanol is:

$n_{e} = \frac{m_{e}}{M_{e}} = \frac{268.13 g}{46.07 g/mol} = 5.82 moles$

And the moles of the solution is:    

$n_{s} = n_{e} + n_{w}$

Where w is for water

$n_{s} = n_{e} + \frac{m_{w}}{M_{w}}$

$n_{s} = n_{e} + \frac{m_{s} - m_{e}}{M_{w}}$

$n_{s} = 5.82 moles + \frac{957 g - 268.13 g}{18 g/mol} = 44.09 moles$

Hence, the mole fraction of ethanol is:

$\chi_{e} =\frac{5.82 moles}{44.09 moles}=0.13$

I hope it helps you!