Answer:
a) The freezing point depression = 1.16 °C
b) Psolution = 101.2 torr Psolution/Psolvent = 101.2 / 103 = 0.983
c) osmotic pressure π = 4.85 atm
d) Molecular weight = 41.4 g/mol
Explanation:
Step1 : Data given
Mass of substance = 7.75 grams
Mass of benzene = 825 grams
Boiling point elevation = 0.575 °C
Kf = 5.12 C/m
Kb = 2.53 C/m
Density = 876.6 Kg /m³ = 0.8766 g/mL
Step 2 Calculate the molaltiy
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = 0.575 °C
⇒i = the van't Hoff factor = 1
⇒Kb = the boiling point constant = 2.53 °C/m
⇒with m = TO BE DETERMINED
0.575 = 2.53 * m
m = 0.227 molal
Step 3: Calculate the freezing point depression
ΔT = i*Kf*m
⇒ΔT = the freezing point depressio, = TO BE DETERMINED
⇒i = the van't Hoff factor = 1
⇒Kb = the freezing point constant = 5.12 °C/m
⇒with m = 0.227 molal
ΔT = 5.12 °C/m * 0.227 m
<u>ΔT = 1.16 °C</u>
<u />
Step 4: Calculate the molar mass of the solute
Molality = moles / mass benzene
0.227 molal = moles / 0.825 kg
Moles substance = 0.227 * 0.825
Moles substance = 0.187 moles
Molar mass = mass / moles
Molar mass substance = 7.75 grams / 0.187 moles
<u>Molar mass substance = 41.4 g/mol</u>
<u />
Step 5: Calculate the ratio of the vapor pressure above the solution to that of the pure solvent,
Moles benzene = 825 grams / 78.11 g/mol
Moles benzene = 10.56 moles
Mol fraction benzene = 10.56 moles /(10.56 + 0.187 )
Mol fraction benzene = 0.983
Psolution = 103 * 0.983 = 101.2 atm
Step 5: Calculate moles substance
Moles substance = 7.75 grams / 41.4 g/mol
Moles substance = 0.187 moles
Step7: Calculate volume
Volume = mass / density
Volume = 825 * 10^-3 / 876.6
Volume = 0.942 L
Step 7 Calculate osmotic pressure
π = i*M*R*T = i n/V * R*T
⇒π = the osmotic pressure = TO BE DETERMINED
⇒i = the van't Hoff factor = 1
⇒M = the molar concentration = n/V
⇒R = the gas constant = 0.08206 L*atm/mol*K
⇒T = the temperature = 298 K
<u />
π = 1 * (0.187 moles / 0.942L) * 0.08206 * 298 K
π = 4.85 atm
