Question

A solution is made containing 14.6g of CH3OH in 185g H2O.1. Calculate the mole fraction of CH3OH.2. Calculate the mass percent of CH3OH.3. Calculate the molality of CH3OH.

Answer 1

Answer:

* $x_{CH_3OH}=0.0425$

* $\%m/m_{CH_3OH}=7.31\%$

* $m=2.46m$

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

$x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}$

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

$n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O$

Hence, mole fraction is:

$x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425$

Next, mass percent is:

$\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%$

And the molality, considering the mass of water in kg (0.185 kg):

$m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m$

Regards.