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:
[tex]\%_{m/m} = \frac{m_{e}}{m_{s}} \times 100[/tex]
Where:
[tex]m_{e}[/tex] is the mass of ethanol
[tex]m_{s}[/tex] is the mass of the solution
We need to calculate the mass of ethanol and the mass of the solution:
[tex] d = \frac{m_{s}}{V} [/tex]
Where:
d: is the density of the solution = 0.957 g/mL
V: is the volume = 1 L
[tex] m_{s} = d*V = 0.957 \frac{g}{mL}*\frac{1000 mL}{1 L}*1 L = 957 g [/tex]
Now, from the concentration we can find the mass of ethanol:
[tex] C = \frac{n_{e}}{V} = \frac{m_{e}}{M_{e}*V} [/tex]
Where:
[tex]M_{e}[/tex]: is the molar mass of ethanol = 46.07 g/mol
[tex]n_{e}[/tex]: is the number of moles of ethanol = m/M
[tex] m_{e} = C*M*V = 5.82 \frac{mol}{L}*46.07 \frac{g}{mol}*1 L = 268.13 g [/tex]
Finally, the mass percent of ethanol is:
[tex]\%_{m/m} = \frac{268.13 g}{957 g} \times 100 = 28.02 \%[/tex]
2. The mole fraction of ethanol is given by:
[tex] \chi_{e} = \frac{n_{e}}{n_{s}} [/tex]
The number of moles of ethanol is:
[tex]n_{e} = \frac{m_{e}}{M_{e}} = \frac{268.13 g}{46.07 g/mol} = 5.82 moles[/tex]
And the moles of the solution is:
[tex] n_{s} = n_{e} + n_{w} [/tex]
Where w is for water
[tex] n_{s} = n_{e} + \frac{m_{w}}{M_{w}} [/tex]
[tex] n_{s} = n_{e} + \frac{m_{s} - m_{e}}{M_{w}} [/tex]
[tex]n_{s} = 5.82 moles + \frac{957 g - 268.13 g}{18 g/mol} = 44.09 moles[/tex]
Hence, the mole fraction of ethanol is:
[tex]\chi_{e} =\frac{5.82 moles}{44.09 moles}=0.13[/tex]
I hope it helps you!
