Question ea892

Sep 12, 2015

You need to add $\text{442 g}$ 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

$\textcolor{b l u e}{{P}_{i} = {\chi}_{i} \cdot {P}_{i}^{\circ}} \text{ }$, where

${P}_{i}$ - the vapor pressure of the component $i$ in solution;
${\chi}_{i}$ - the mole fraction of component $i$;
${P}_{i}^{\circ}$ - the vapor pressure of pure $i$.

So, the vapor pressure of the solution must be

$P = {P}_{\text{water"^@ - "0.595 mmHg}}$

$P = \text{17.5 mmHg" - "0.595 mmHg" = "16.905 mmHg}$

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

$P = {\chi}_{\text{water" * P_"water"^@ implies chi_"water" = P/P_"water}}^{\circ}$

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

${\chi}_{\text{water" = (16.905color(red)(cancel(color(black)("mmHg"))))/(17.5color(red)(cancel(color(black)("mmHg")))) = "0.966}}$

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.

${\chi}_{\text{water" = n_"water"/n_"total}}$

The total number of moles in the solution will be

${n}_{\text{total" = n_"water" + n_"sucrose}}$

This means that you have

${\chi}_{\text{water" = n_"water"/(n_"water" + n_"sucrose") implies n_"sucrose" = n_"water"/chi_"water" - n_"water}}$

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

${n}_{\text{sucrose" = 36.692/0.966 - 36.692 = "1.291 moles sucrose}}$

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")#