Question

a soluation of acetone in water has a molarity of 2.422M and a density of 0.970 g/mL. Calculate the mole fraction

Answer 1

`x_(acetone)=7.970x10^(-3)`

Step-by-step 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:

`n=0.422mol/L*1L=0.422mol`

Then, with the density of solution, we can compute the mass of the solution for the selected 1-L volume basis:

`m_(solution)=1L*(1000mL)/(1L)*(0.970g)/(1mL)=970g`

After that, we compute the mass of water in the solution, considering the mass of acetone (molar mass = 58.08 g/mol):

`m_(H_2O)=970g-0.422molAcetone*(58.08g\ Acetone)/(1mol\ Acetone) =945.49gH_2O`

Next, the moles of water:

`n_(H_2O)=945.49g*(1molH_2O)/(18gH_2O) =52.53molH_2O`

Finally, the mole fraction:

`x_(acetone)=(n_(acetone))/(n_(acetone)+n_(H_2O))=(0.422mol)/(0.422mol+52.53mol)\\ \\x_(acetone)=7.970x10^(-3)`

Regards.