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ure to answer all parts. The vapor pressure of ethanol (C2H5OH) at 20°C is 44 mmHg, and the vapor pressure of methanol (CH3OH) at the same temperature is 94 mmHg. A mixture of 25.3 g of methanol and 47.1 g of ethanol is prepared and can be assumed to behave as an ideal solution. Calculate the vapor pressure of methanol and ethanol above this solution at 20°C. Be sure to report your answers to the correct number of significant figures.

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Answer:

[tex]P_{met}=40.98mmHg\\\\P_{et}=24.82mmHg[/tex]

Explanation:

Hello

In this case, by considering the Raoult's law we can write:

[tex]P_{met}=x_{met}P_{sat,met}\\\\P_{et}=x_{et}P_{sat,et}[/tex]

Whereas the purpose is to compute the pressures of both methanol and ethanol (pressures above the solution), thus, the first step is to compute the molar fractions:

[tex]n_{met}=25.3gCH_3OH*\frac{1molCH_3OH}{32gCH_3OH}=0.791molCH_3OH\\ \\n_{et}=47.1gCH_3CH_2OH*\frac{1molCH_3CH_2OH}{46gCH_3CH_2OH}=1.02molCH_3CH_2OH\\\\x_{met}=\frac{0.791mol}{0.791mol+1.02mol}=0.436\\ \\x_{et}=1-x_{met}=1-0.436=0.564[/tex]

Therefore, the pressures turns out:

[tex]P_{met}=0.436*94mmHg=40.98mmHg\\\\P_{et}=0.564*44mmHg=24.82mmHg[/tex]

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