Question

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 ? %.

Answer 1

Answer:

6%

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:

$m_{solution}=1L*\frac{1000 mL}{1L}*\frac{0.988g}{1mL} =988g$

Now, the mass of the ethanol:

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

Finally, the by mass percent is:

$m/m=\frac{59.34g}{988g}*100$%

%m=6%

Best regards.