Question

Question #ea892

Answer 1

You need to add `"442 g"` of sucrose.

Explanation:

The idea here is that you need to figure out what mole fraction of water is needed in order for the solution to have a vapor pressure 0.595 mmHg smaller than that of pure water.

You know that the vapor pressure for a volatile component of a solution is

`color(blue)(P_i = chi_i * P_i^@)" "`, where

`P_i` - the vapor pressure of the component `i` in solution;
`chi_i` - the mole fraction of component `i`;
`P_i^@` - the vapor pressure of pure `i`.

So, the vapor pressure of the solution must be

`P = P_"water"^@ - "0.595 mmHg"`

`P = "17.5 mmHg" - "0.595 mmHg" = "16.905 mmHg"`

Since water is the only volatile component, you can say that

`P = chi_"water" * P_"water"^@ implies chi_"water" = P/P_"water"^@`

This means that the mole fraction of water in the solution must be

`chi_"water" = (16.905color(red)(cancel(color(black)("mmHg"))))/(17.5color(red)(cancel(color(black)("mmHg")))) = "0.966"`

The mole fraction of water is simply the ratio between the number of moles of water and the total number of moles present in the solution.

`chi_"water" = n_"water"/n_"total"`

The total number of moles in the solution will be

`n_"total" = n_"water" + n_"sucrose"`

This means that you have

`chi_"water" = n_"water"/(n_"water" + n_"sucrose") implies n_"sucrose" = n_"water"/chi_"water" - n_"water"`

Use water's molar mass to find the number of moles of water

`661color(red)(cancel(color(black)("g"))) * "1 mole"/(18.015color(red)(cancel(color(black)("g")))) = "36.692 moles"`

This means that the solution must contain

`n_"sucrose" = 36.692/0.966 - 36.692 = "1.291 moles sucrose"`

Now use sucrose's molar mass to see how many grams would contain this many moles

`1.291color(red)(cancel(color(black)("moles"))) * "342.29 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("442 g")`