Answer to Question #189725 in General Chemistry for Gwyneth Balete

Question

what is the vapor pressure of a solution made by dissolving 225 grams of glucose in 358 mL of water at 30°C? The molar mass of glucose in 180.2 g/mol. What is the vapor pressure lowering? The vapor pressure of pure water at 30°C is 31.82 mmHg. Assume that density of the solution is 1.00 g/mL.

Answer 1

Q189725

what is the vapor pressure of a solution made by dissolving 225 grams of glucose in 358 mL of water at 30°C? The molar mass of glucose is 180.2 g/mol. What is the vapor pressure lowering? The vapor pressure of pure water at 30°C is 31.82 mmHg. Assume that the density of the solution is 1.00 g/mL.

Solution :

Vapor pressure lowering is related to the mole fraction of solute in the solution by the formula.

$\Delta{P} _{solvent} = \chi_{solute} * P_{solvent}$

χ _solute is the mole fraction of solute.

ΔP_solvent is the vapor pressure lowering.

P_solvent is the vapor pressure of the solvent.

First find the moles of glucose present in 225 grams.

$$moles \ of glucose = 225 \cancel{\ g \ of \ glucose } * \frac{1 \ mol \ of \ glucose }{180.2 \cancel{ \ g \ of \ glucose } }$$

$$= 1.2486 \ mol \ of \ glucose$$

Next, find the moles of water in 358 mL. The density of water is 1.00g/mL.

mass of water = density * volume = 1.00 g/mL * 358 mL

= 358 grams.

$$moles \ of H_2O= 358 \cancel{\ g \ of \ H_2O } * \frac{1 \ mol \ of \ H_2O }{18.015 \cancel{ \ g \ of \ H_2O } }$$

$$= 19.872 \ moles \ of \ H_2O$$

$$moles \ fraction \ of \ glucose, \ \chi_{glucose} = \frac{moles \ of \ glucose }{moles \ of \ glucose \ + \ moles \ of \ H_2O}$$

$$\ \chi_{glucose} = \frac{1.2486 \ moles \ of \ glucose }{1.2486 \ moles \ of \ glucose \ + \ 19.872 \ moles \ of \ H_2O}$$

$$\ \chi_{glucose} = \frac{1.2486 }{21.1209} = 0.05912 ;$$

Now we know,

χ _glucose is = 0.05912 , and P water = 31.82 mm Hg.

glucose is the solute and water is the solvent here.

$$\Delta{P} _{water } = \chi_{glucose } * P_{water}$$

$$\Delta{P} _{water } = 0.05912 \ * \ 31.82 \ mm \ Hg.$$

$$\Delta{P} _{water } = 1.881 \ mm \ Hg ;$$

In the correct significant figure, the answer is 1.88 mm Hg.

Hence the vapor pressure of water is lowered by 1.88 mm Hg.