# Question eb24b

##### 1 Answer
Feb 11, 2016

${\text{100 g mol}}^{- 1}$

#### Explanation:

!! LONG ANSWER !!

The first thing to note here is that the vapor pressure of a solution is lower than the vapor pressure of the pure solvent. Keep this in mind.

Now, according to Raoult's Law, when dealing with a non-volatile solute, the vapor pressure of the solution will depend on the mole fraction of the solvent, which we'll label as $A$, and on the vapor pressure of the pure solvent

color(blue)(P_"sol" = chi_A xx P_A^@)" ", where

${P}_{\text{sol}}$ - the vapor pressure of the solution
${\chi}_{A}$ - the mole fraction of the solvent
${P}_{A}^{\circ}$ - the vapor pressure of the pure solvent

You can express the mole fraction of the solvent in terms of the mole fraction of the solute, which we'll label as $B$, since your solution only contains a solute and a solvent

$\textcolor{b l u e}{{\chi}_{A} + {\chi}_{B} = 1}$

${\chi}_{A} = 1 - {\chi}_{B}$

Plug this into the above equation to get

${P}_{\text{sol}} = \left(1 - {\chi}_{B}\right) \cdot {P}_{A}^{\circ}$

This is equivalent to

${P}_{\text{sol}} = {P}_{A}^{\circ} - {\chi}_{B} \cdot {P}_{A}^{\circ}$

Rearrange to isolate the mole fraction of the solute on one side of the equation

${\chi}_{B} \cdot {P}_{A}^{\circ} = {P}_{A}^{\circ} - {P}_{\text{sol}}$

${\chi}_{B} = \frac{{P}_{A}^{\circ} - {P}_{\text{sol}}}{P} _ {A}^{\circ}$

Now, the mole fraction of the solute is defined as the number of moles of solute, say ${n}_{B}$, divided by the total number of moles present in solution, which is ${n}_{B} + {n}_{A}$.

This means that you have

$\frac{{P}_{A}^{\circ} - {P}_{\text{sol}}}{P} _ {A}^{\circ} = {n}_{B} / \left({n}_{B} + {n}_{A}\right)$

This is equivalent to

${P}_{A}^{\circ} / \left({P}_{A}^{\circ} - {P}_{\text{sol}}\right) = \frac{{n}_{B} + {n}_{A}}{n} _ B$

${P}_{A}^{\circ} / \left({P}_{A}^{\circ} - {P}_{\text{sol}}\right) = 1 + {n}_{A} / {n}_{B}$

Rearrange to get

${P}_{A}^{\circ} / \left({P}_{A}^{\circ} - {P}_{\text{sol}}\right) - 1 = {n}_{A} / {n}_{B}$

$\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{A}^{\circ}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{A}^{\circ}}}} + {P}_{\text{sol")/(P_A^@ - P_"sol}}\right) = {n}_{A} / {n}_{B}$

P_"sol"/(P_A^@ - P_"sol") = n_A/n_B

But you know that you can write the number of moles as the ratio between the mass of the substance, $m$, and its molar mass, $M$

${n}_{A} = {m}_{A} / {M}_{M A} \text{ }$ and $\text{ } {n}_{B} = {m}_{B} / {M}_{M B}$

Plug this into the above equation to get

P_"sol"/(P_A^@ - P_"sol") = m_A/M_(MA) * M_(MB)/m_B

Finally, rearrange to get

$\textcolor{b l u e}{\left({P}_{A}^{\circ} - {P}_{\text{sol")/P_"sol" = m_B/m_A * M_(MA)/M_(MB))" " " "color(red)(("*}}\right)}$

Here

${m}_{a}$, ${M}_{M A}$ - the mass and molar mass of the solvent
${m}_{b}$, ${M}_{M B}$ - the mass and molar mass of the solute

Now, the most important thing to realize now is that the loss in weight for the solution and the pure solvent is proportional to their respective vapor pressures.

Here's why.

When dry air is passed through the solution, solvent molecules will pass into the dry air. The dry air will be saturated proportional to the vapor pressure of the solution.

$\Delta {m}_{\text{sol" prop P_"sol}}$

Here $\Delta {m}_{\text{sol}}$ represents the change in the mass of the solution.

The air is then passed through the pure solvent. Since the vapor pressure of the pure solvent is higher than that of the solution, the air will absorb additional water molecules.

More specifically, the air will absorb water molecules proportional to the difference between the vapor pressure of the solution and the vapor pressure of the solvent, which we've labeled as $A$

$\Delta {m}_{A} \propto \left({P}_{A}^{\circ} - {P}_{\text{sol}}\right)$

Here $\Delta {m}_{A}$ represents the change in the mass of the pure solvent.

Plug this into equation $\textcolor{red}{\left(\text{*}\right)}$ to get

$\textcolor{b l u e}{\frac{\Delta {m}_{A}}{\Delta {m}_{\text{sol}}} = {m}_{B} / {m}_{A} \cdot {M}_{M A} / {M}_{M B}}$

Now you have all that you need to solve for ${M}_{M B}$, the molar mass of the solute. Water's molar mass is listed as ${\text{18.015 g mol}}^{- 1}$. Rearrange to get

${M}_{M B} = {m}_{B} / {m}_{A} \cdot \frac{\Delta {m}_{\text{sol}}}{\Delta {m}_{A}} \cdot {M}_{M A}$

M_(MB) = (10color(red)(cancel(color(black)("g"))))/(90color(red)(cancel(color(black)("g")))) * (2.5color(red)(cancel(color(black)("g"))))/(0.05color(red)(cancel(color(black)("g")))) * "18.015 g mol"^(-1)#

${M}_{M B} = {\text{100.0833 g mol}}^{- 1}$

Rounded to one sig fig, the answer will be

${M}_{M B} = \textcolor{g r e e n}{{\text{100 g mol}}^{- 1}}$