The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. determine the value of the van't hoff factor for this solute if the kb for water is 0.512°c/m. 2.7 1.8 2.3 3.0 3.6

The value of the van't Hoff factor for the given solute is 2.7

Option (A) 2.7 is correct.

The Van't Hoff factor is a measure of a solute's effect on colligative qualities such boiling point elevation, osmotic pressure, and relative vapor pressure reduction.

Given,

Molar mass is 132.15 g/mol

Boiling point elevation is 102.5 °C

kb for water is  0.512°c/m

Molarity of solution is 1.83 m

The value of van't Hoff factor is: 2.66 or 2.7 (approx)

Now, from the solution of colligative properties to calculate elevation in boiling point.

where Δ = elevation of boiling point (102.5)

Thus, The value is 2.7 option (A) is correct.

Learn more about van't Hoff factor, here:

link

We need to know the value of van't hoff factor.

The van't hoff factor is: 2.66 or 2.7 (approximately)

(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.

From the equation: Δ=i .m, where Δ= elevation of boiling point=102.5 - 100=2.5°C.

m=molality of solute=1.83 m (Given)

= Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)

i= Van't Hoff factor

So, 2.5= i X 0.512 X 1.83

i=

i=2.66= 2.7 (approx.)

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.

the answer is b & c i believe : )