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lions [1.4K]
3 years ago
7

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

? %.
Chemistry
1 answer:
anzhelika [568]3 years ago
4 0

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.

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