Answer
(d) Mole % = 9.02%
(e) Mass % = 15.0%
Explanation
Given that:
% methanol by mass of the aqueous solution = 15.0%
Density of methanol = 0.998 g/mol
What to find:
To calculate (d) Mole % (e) Mass %
Step-step solution:
(d) Mole %
You need to pick a sample volume of this solution and use the given density to find its mass.
Let's take a 1.00 L sample of this solution. The solution has a density of
0.998 g/mL, meaning that you get 0.998 g for every 1 mL of solution.
The mass of the 1.00 L sample will thus be:
[tex]\begin{gathered} Mass=Density\times Volume=0.998g\text{/}mL\times(1.00\times1000mL) \\ \\ Mass=998\text{ }g \end{gathered}[/tex]Now, it is known that this solution is 15.0% w/w methanol. In simple terms, this tells you that you get 15.0 g of ethanol for every 100 g of solution.
So the sample will contain:
[tex]998g\text{ }solution\times\frac{15g\text{ }methanol}{100g\text{ }solution}=149.7g\text{ }methanol[/tex]This implies that the solution will also contain:
[tex]\begin{gathered} Mass\text{ }of\text{ }solution=Mass\text{ }of\text{ }water-mass\text{ }of\text{ }methanol \\ \\ 998g=Mass\text{ }of\text{ }water-149.7g \\ \\ Mass\text{ }of\text{ }water=998g-149.7g=848.30g \end{gathered}[/tex]To get the mole % of methanol, you must determine how many moles of methanol and of water the solution contains.
[tex]\begin{gathered} Moles\text{ }of\text{ }methanol=\frac{mass}{molar\text{ }mass}=\frac{149.7g}{32.04g\text{/}mol}=4.67\text{ }mol \\ \\ Moles\text{ }of\text{ }water=\frac{848.30g}{18.02g\text{/}mol}=47.08\text{ }mol \end{gathered}[/tex]Hence, the mole % of methanol will be:
[tex]\begin{gathered} Mole\text{ }\%\text{ }methanol=\frac{moles\text{ }of\text{ }methanol}{Total\text{ }mole}\times100\%=\frac{4.67\text{ }mol}{\placeholder{⬚}} \\ \\ Mole\text{ }\%\text{ }methanol=\frac{4.67\text{ }mol}{4.67\text{ }mol+47.08\text{ }mol}\times100\%=9.02\% \end{gathered}[/tex]Hence, the mole % methanol of the aqueous solution is 9.02%.
(e) Mass %
The mass % of methanol in the aqueous solution has been given in the question to be 15.0%.
Also, you can calculate it from the masses of methanol and water above as follows:
[tex]\begin{gathered} Mass\text{ }\%\text{ }methanol=\frac{mass\text{ }of\text{ }methanol}{Total\text{ }mass}\times100\% \\ \\ Mass\text{ }\%\text{ }methanol=\frac{149.7g}{998g}\times100\%=15.0\% \end{gathered}[/tex]
