An aqueous solution of 1.29 M ethanol, CH3CH2OH, has a density of 0.988 g/mL. The percent by mass of CH3CH2OH in the solution is ? %.

Question

asked Sep 10, 2020 48.5k views

1 Answer

Eric HB · Answer 1 · 2020-09-16T02:05:42+0000

Answer:

6%

Step-by-step explanation:

Hello, for this case, we assume that the volume of the solution is 1L, thus, the mass is given by using the density as follows:

m_(solution)=1L*(1000 mL)/(1L)*(0.988g)/(1mL) =988g

Now, the mass of the ethanol:

$m_(C_2H_5OH)=(1.29molC_2H_5OH/L*1L)*(46gC_2H_5OH)/(1molC_2H_5OH) \\m_(C_2H_5OH)=59.34g$

Finally, the by mass percent is:

$m/m=(59.34g)/(988g)*100\\$ %

%m=6%

Best regards.