Answer:
[tex]x_{acetone}=7.970x10^{-3}[/tex]
Explanation:
Hello,
In this case, for the given molarity, we can assume a volume of 1 L of solution, to obtain the following moles of acetone:
[tex]n=0.422mol/L*1L=0.422mol[/tex]
Then, with the density of solution, we can compute the mass of the solution for the selected 1-L volume basis:
[tex]m_{solution}=1L*\frac{1000mL}{1L}*\frac{0.970g}{1mL}=970g[/tex]
After that, we compute the mass of water in the solution, considering the mass of acetone (molar mass = 58.08 g/mol):
[tex]m_{H_2O}=970g-0.422molAcetone*\frac{58.08g\ Acetone}{1mol\ Acetone} =945.49gH_2O[/tex]
Next, the moles of water:
[tex]n_{H_2O}=945.49g*\frac{1molH_2O}{18gH_2O} =52.53molH_2O[/tex]
Finally, the mole fraction:
[tex]x_{acetone}=\frac{n_{acetone}}{n_{acetone}+n_{H_2O}}=\frac{0.422mol}{0.422mol+52.53mol}\\ \\x_{acetone}=7.970x10^{-3}[/tex]
Regards.
