Answer:
* [tex]x_{CH_3OH}=0.0425[/tex]
* [tex]\%m/m_{CH_3OH}=7.31\%[/tex]
* [tex]m=2.46m[/tex]
Explanation:
Hello,
In this case, for the mole fraction of methanol we use the formula:
[tex]x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}[/tex]
Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):
[tex]n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O[/tex]
Hence, mole fraction is:
[tex]x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425[/tex]
Next, mass percent is:
[tex]\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%[/tex]
And the molality, considering the mass of water in kg (0.185 kg):
[tex]m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m[/tex]
Regards.
