# Question #eb24b

##### 1 Answer

#### 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 *pure solvent*

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

You can express the mole fraction of the solvent in terms of the mole fraction of the solute, which we'll label as

#color(blue)(chi_A + chi_B = 1)#

#chi_A = 1 - chi_B#

Plug this into the above equation to get

#P_"sol" = (1 - chi_B) * P_A^@#

This is equivalent to

#P_"sol" = P_A^@ - chi_B * P_A^@#

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

#chi_B * P_A^@ = P_A^@ - P_"sol"#

#chi_B = (P_A^@ - P_"sol")/P_A^@#

Now, the mole fraction of the solute is defined as the number of moles of solute, say **total number of moles** present in solution, which is

This means that you have

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

This is equivalent to

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

#P_A^@/(P_A^@ - P_"sol") = 1 + n_A/n_B#

Rearrange to get

#P_A^@/(P_A^@ - P_"sol") -1 = n_A/n_B#

#(color(red)(cancel(color(black)(P_A^@))) - color(red)(cancel(color(black)(P_A^@))) + P_"sol")/(P_A^@ - P_"sol") = 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, **molar mass**,

#n_A = m_A/M_(MA)" "# and#" "n_B = m_B/M_(MB)#

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

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

Here

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.

#Deltam_"sol" prop P_"sol"#

Here

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

#Deltam_A prop (P_A^@ - P_"sol")#

Here

Plug this into equation

#color(blue)((Deltam_A)/(Deltam_"sol") = m_B/m_A * M_(MA)/M_(MB))#

Now you have all that you need to solve for

#M_(MB) = m_B/m_A * (Deltam_"sol")/(Deltam_A) * M_(MA)#

#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_(MB) = "100.0833 g mol"^(-1)#

Rounded to one **sig fig**, the answer will be

#M_(MB) = color(green)("100 g mol"^(-1))#