Part #1: 30.0 g of glucose (C6H12O6, molar mass = 180.16 g/mol) are dissolved in
ID: 913517 • Letter: P
Question
Part #1:
30.0 g of glucose (C6H12O6, molar mass = 180.16 g/mol) are dissolved in 100. grams of water (molar mass 18.02). What is the mole fraction of water in solution? Note: glucose is a non-volatile solute.
Part #2:
At 25.0 oC, pure water has a vapor pressure of 24.2 torr. What would the vapor pressure of the solution in part one be (in torr)?
Part #3:
A liquid is a mixture of benzene and toluene. The sample contains 0.225 moles of benzene and 0.775 moles of toluene. At a given temperature, pure benzene has a vapor pressure of 38.02 torr, while pure toluene has a vapor pressure of 8.73 torr. What is the total combined vapor pressure of the two liquids (in torr?)?
Explanation / Answer
1)
mass of glucose = 30.0 g
Molar mass of glucose = 180.16 g/mol
number of moles of glucose = mass / molar mass
= 30.0 / 180.16
= 0.1665 mol
mass of solvent = 100 g
Molar mass of water = 18 g/ mol
number of moles of water = mass / molar mass
= 100/18
= 5.56 mol
Mole fraction of water = mol of water / total moles
= 5.56 / (5.56 + 0.1665)
= 0.97
Answer: 0.97
2)
Let pressure be P
usE:
P = PO*mole fraction
P = 24.2 * 0.97
= 23.5 torr
Answer: 23.5 torr
3)
mole fraction (benzene) = mol of benzene/ total moles
=0.225 / (0.225 + 0.775)
=0.225
mole fraction (toluene) = mol of toluene/ total moles
=0.775 / (0.225 + 0.775)
=0.775
P total = Po benzene * mole fraction (benzene) + Po toluene * mole fraction (toluene)
= 38.02*0.225 + 8.73*0.775
= 15.32 torr
Answer: 15.32 torr