Question

When 1 mol of a nonvolatile, nondissociating solute is dissolved in 3 mol of volatile solvent, the ratio of vapor pressure of the solution to that of the pure solvent (at the same temperature) is approximately:________

Answer 1

`(P_(solution))/(P_(solvent)^(vap)) =0.75`

Step-by-step explanation:

Hello there!

In this case, since the solvation of a nonvolatile-nondissociating solute in a volatile solvent is modelled via the Raoult's law:

`P_(solution)=x_(solvent)P_(solvent)^(vap)`

Thus, we can calculate the ratio of the vapor pressure of the solution to that of the pure solvent, mole fraction, as shown below:

`x_(solvent)=(P_(solution))/(P_(solvent)^(vap)) =(n_(solvent))/(n_(solute)+n_(solvent))`

Thus, we plug in the moles of solvent and solute to obtain:

`(P_(solution))/(P_(solvent)^(vap)) =(3)/(3+1)\\\\ (P_(solution))/(P_(solvent)^(vap)) =0.75`

Regards!