91Ó°ÊÓ

Mole ratio of ethanol and methanol in liquid can be calculated by dividing moles of ethanol in vapor and partial vapor pressure of ethanol and moles of methanol divided by partial vapor pressure of methanol.

The mixture contains a $\raisebox{1ex}{$97$}\!\left/ \!\raisebox{-1ex}{$1$}\right.$mass ratio of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$and ${\text{CH}}_{\text{3}}\text{OH}$.

Hence,

The mass of ethanol is $97\text{g}$.

The mass of methanol is $\text{1g}$.

Calculate the mass of ethanol as follow.

$\begin{array}{rcl}\text{³¾´Ç±ô²¹°ù″¾²¹²õ²õ​â¶Ä‰o´Ú e³Ù³ó²¹²Ô´Ç±ô}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)& =& 2×12+1×6+16\\ & =& 46\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

Calculate the mass of methanol as follow.

$\begin{array}{rcl}\text{³¾´Ç±ô²¹°ù″¾²¹²õ²õ​â¶Ä‰o´Ú″¾±ð³Ù³ó²¹²Ô´Ç±ô}\left({\text{CH}}_{\text{3}}\text{OH}\right)& =& 12+1×4+16\\ & =& 32\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

Moles of ethanol can be calculated as follow.

$\begin{array}{rcl}\text{³¾´Ç±ô±ð²õ​â¶Ä‰o´Ú″¾±ð³Ù³ó²¹²Ô´Ç±ô}\left({\text{CH}}_{\text{3}}\text{OH}\right)& =& \text{1 g}×\frac{\text{1″¾´Ç±ô″¾±ð³Ù³ó²¹²Ô´Ç±ô}}{\text{32 }{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 0.031\text{mol}\end{array}$

Moles of methanol can be calculated as follow.

$\begin{array}{rcl}\text{³¾´Ç±ô±ð²õ​â¶Ä‰o´Ú e³Ù³ó²¹²Ô´Ç±ô}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)& =& \text{97g}×\frac{\text{1″¾´Ç±ô″¾±ð³Ù³ó²¹²Ô´Ç±ô}}{\text{46 }{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 2.11\text{mol}\end{array}$

Given the partial vapor pressure of ethanol and methanol is $60.5\text{torr}$androle="math" localid="1663511569425" $\text{126torr}$ respectively.

Calculate the mole ratio of ethanol and methanol in liquid as below.

$\begin{array}{rcl}\frac{\text{³¾´Ç±ô±ð²õ o´Ú e³Ù³ó²¹²Ô´Ç±ô​â¶Ä‰i²Ô l¾±±ç³Ü¾±»å}}{\text{³¾´Ç±ô±ð²õ o´Ú″¾±ð³Ù³ó²¹²Ô´Ç±ô i²Ô l¾±±ç³Ü¾±»å}}& =& \frac{{\displaystyle \left(\frac{\text{³¾´Ç±ô±ð²õ o´Ú e³Ù³ó²¹²Ô´Ç±ô i²Ô v²¹±è´Ç°ù}}{\text{±è²¹°ù³Ù¾±²¹±ô v²¹±è´Ç°ù p°ù±ð²õ²õ³Ü°ù±ð o´Ú e³Ù³ó²¹²Ô´Ç±ô}}\right)}}{{\displaystyle \left(\frac{\text{³¾´Ç±ô±ð²õ o´Ú″¾±ð³Ù³ó²¹²Ô´Ç±ô i²Ô v²¹±è´Ç°ù}}{\text{±è²¹°ù³Ù¾±²¹±ô v²¹±è´Ç°ù p°ù±ð²õ²õ³Ü°ù±ð o´Ú″¾±ð³Ù³ó²¹²Ô´Ç±ô}}\right)}}\\ & =& \frac{{\displaystyle \left(\frac{\text{2.11}}{\text{60.5torr}}\right)}}{{\displaystyle \left(\frac{\text{0.031}}{\text{126torr}}\right)}}\\ & =& \frac{142}{1}\end{array}$

It knows the molar mass of ethanol and methanol are $46.7\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mo}$}\right.$and $32.04\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$respectively.

Since it can calculate the mass of any substance by multiplying its moles and molar mass.

Hence, you can calculate the mass ratio of ethanol and methanol by multiplying their moles with their molar mass as follow.

$\begin{array}{rcl}\frac{\text{³¾²¹²õ²õ o´Ú e³Ù³ó²¹²Ô´Ç±ô​â¶Ä‰i²Ô l¾±±ç³Ü¾±»å}}{\text{³¾²¹²õ²õ o´Ú e³Ù³ó²¹²Ô´Ç±ô​â¶Ä‰i²Ô l¾±±ç³Ü¾±»å}}& =& \frac{142\text{″¾´Ç±ô}×46.07{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}{1\text{mol}×32.04{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& \frac{204}{1}\end{array}$

Hence, the mass ratio of ethanol and methanol in liquid is $\raisebox{1ex}{${\displaystyle 204}$}\!\left/ \!\raisebox{-1ex}{$1$}\right.$.