Chemistry help asap😭

Question

What is the freezing point of a 0.20 m sucrose (nonelectrolyte)? Kf = 1.86 o C /m.

Calculate the osmotic pressure of 0.2 M salt solution (NaCl) at 25 o C?

Answer 1

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 1 for sucrose, a non electrolyte

m = molality = 0.2

K = freezing constant = 1.86º/m

Solving for ∆T we have ...

∆T = (1)(0.2)(1.86) = 0.37º

Freezing point will be -0.37ºC

π = iMRT

π = osmotic pressure = ?

i = van't Hoff factor = 2 for NaCl as it dissociates into 2 ions (Na⁺ and Cl^-)

M = molarity = 0.2

R = gas constant = 0.0821 Latm/Kmol)

T = temperature in K = 25º + 273 = 298K

Solving for π, we have ...

π = (2)(0.2)(0.0821)(298)

π = 9.8 atm = 10 atm (1 sig. fig.)

Answer 2

The freezing point is depressed by the presence of the solute. It is a nonelectrolyte, so, for water:

ΔT_FP = k_fpd x molality of solute = 1.86 °C/molal * 0.2 molal sucrose. subtract this from 0°C and you have the new freezing point.

Osmotic pressure, π = CRT with respect to water. That is that the pure water would push into the concentrated solution with that pressure. The osmotic pressure between two concentrated solution would be based on the difference in concentration.

π = .2 mole/l * 0.0821 liter*atm/mole*K * 298K for π in atmospheres.

You might think that the MKS R (8.314 J/mole*K) but there is an issues with the liters being the wrong units. If you recognize that the Pa*m³ is a kPa*liter, than you see that the 8.314 R results in kPa)