For which of the following aqueous solutions would one expect to have the largest van’t hoff factor (i)? a. 0.050 m nacl b. 0.500 m k2so4 c. 0.050 m k2so4 d. 0.500 m nacl e. 0.500 m c6h12o6 (glucose)

The van't hoff factor of 0.500m K₂SO₄ will be highest.

Explanation:

Van't Hoff factor was introduced for better understanding of colligative property of a solution.

By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.

a) For NaCl the van't Hoff factor is 2

b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]

Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.

c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.

answer

the ions in solute supperate from eachouthere

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm