Answer:
[tex]0.86M[/tex]
Explanation:
Hello,
In this case, one must consider that the ethanol's concentration in beer is defined as:
[tex]\% et =\frac{m_{et}}{m_{beer}}[/tex]
In such a way, since molarity is defined as:
[tex]M=\frac{mol_{et}}{Volume_{solution}}[/tex]
One is asked to compute the moles, considering a basis of 5 milliliters of ethanol and 100 milliliters of beer, thus:
[tex]mol_{et}=5mL*\frac{0.789g}{1mL} *\frac{1mol}{46g}=0.086mol\ et[/tex]
Now, the volume in molarity is required in liters, therefore:
[tex]Volume_{beer}=100mL*\frac{1L}{1000mL}=0.1L[/tex]
Therefore, the molarity turns out:
[tex]M=\frac{0.086mol}{0.1L}=0.86M[/tex]
Best regards.
