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[tex]T_{b}[/tex]=i [tex]K_{b}[/tex].m, where Δ[tex]T_{b}[/tex]= elevation of boiling point=102.5 - 100=2.5°C.
m=molality of solute=1.83 m (Given)
[tex]K_{b}[/tex]= 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=[tex]\frac{2.5}{0.512 X 1.83}[/tex]
i=2.66= 2.7 (approx.)
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.
[tex]\bold{\Delta Tb = ikbm}[/tex]
where Δ = elevation of boiling point (102.5)
[tex]\bold{102.5\times i\times 0.512 \times 1.83 }[/tex]
[tex]\bold{i = 2.66 = 2.7 }[/tex]
Thus, The value is 2.7 option (A) is correct.
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