Question #ea892

1 Answer
Sep 12, 2015

Answer:

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")#