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Levart [38]
3 years ago
8

A 0.050 M solution of AlCl3 had an observed osmotic pressure of 3.85 atmatm at 20°C.Calculate the van't Hoff factor iii for AlCl

3.
Chemistry
1 answer:
Alja [10]3 years ago
6 0

Answer:

The actual Van't Hoff factor for AlCl3 is 3.20

Explanation:

Step 1: Data given

Molarity of AlCl3 = 0.050 M

osmotic pressure = 3.85 atm

Temperature = 20 °C

Step 2: Calculate the Van't Hoff factor

AlCl3(aq) → Al^3+(aq) + 3Cl^-(aq)

The theoretical value is 4 ( because 1 Al^3+ ion + 3 Cl- ions) BUT due to the interionic atractions the actual value will be less

Osmotic pressure depends on the molar concentration of the solute but not on its identity., and is calculated by:

π = i.M.R.T

 ⇒ with π = the osmotic pressure = 3.85 atm

⇒ with i = the van't Hoff factor

⇒ with M = the molar concentration of the solution = 0.050 M

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 20 °C = 293.15 Kelvin

i = π /(M*R*T )

i = (3.85) / (0.050*0.08206*293.15)

i = 3.20

The actual Van't Hoff factor is 3.20

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