Question

A solution is prepared by condensing 4.00 L of a gas,

Answer 1

Answer:

-2.3 ºC

Explanation:

Kf (benzene) = 5.12 ° C kg mol – 1

1st - We calculate the moles of condensed gas using the ideal gas equation:

n = PV / (RT)

P = 748/760 = 0.984 atm

T = 270 + 273.15 = 543.15 K

V = 4 L

R = 0.082 atm.L / mol.K

n = (0.984atm * 4L) / (0.082atm.L / K.mol * 543.15K) = 0.088 mol

Then, you calculate the molality of the solution:

m = n / kg solvent

m = 0.088 mol / 0.058 kg = 1.52mol / kg

Then you calculate the decrease in freezing point (DT)

DT = m * Kf

DT = 1.52 * 5.12 = 7.8 ° C

Knowing that the freezing point of pure benzene is 5.5 ºC, we calculate the freezing point of the solution:

T = 5.5 - 7.8 = -2.3 ºC