The molar mass of the solute sodium hydroxide obtained from the measurement of the osmotic pressure of its aqueous solution at 27° C is 25 g mol -1. Therefore, its ionization percentage in this solution is

Question

The molar mass of the solute sodium hydroxide obtained from the measurement of the osmotic pressure of its aqueous solution at 27° C is 25 g mol -1. Therefore, its ionization percentage in this solution is (A) 75 (B) 60 (C) 80 (D) 70

Tardigrade Admin · Accepted Answer

van't Hoff factor, i=( text observed molar mass / text calculated molar mass ) =(40/25)=1.6 Since, sodium hydroxide is binary electrolyte, i=[1+(2-1) α](α= degree of ionisation) 1.6=(1+α) α =1.6-1=0.6=60

Q. The molar mass of the solute sodium hydroxide obtained from the measurement of the osmotic pressure of its aqueous solution at $2 7^{\circ} C$ is $25 g m o l^{- 1}$ . Therefore, its ionization percentage in this solution is

75

60

80

70

van't Hoff factor,
$i = \frac{observed molar mass}{calculated molar mass}$
$= \frac{40}{25} = 1.6$
Since, sodium hydroxide is binary electrolyte,
$i = [1 + (2 - 1) α] (α =$ degree of ionisation)
$1.6 = (1 + α) α$
$= 1.6 - 1 = 0.6 = 60$

Q. The molar mass of the solute sodium hydroxide obtained from the measurement of the osmotic pressure of its aqueous solution at 27∘C is 25gmol−1. Therefore, its ionization percentage in this solution is

75

60

80

70

van't Hoff factor, i= calculated molar mass observed molar mass ​ =2540​=1.6 Since, sodium hydroxide is binary electrolyte, i=[1+(2−1)α](α= degree of ionisation) 1.6=(1+α)α =1.6−1=0.6=60

Q. The molar mass of the solute sodium hydroxide obtained from the measurement of the osmotic pressure of its aqueous solution at $2 7^{\circ} C$ is $25 g m o l^{- 1}$ . Therefore, its ionization percentage in this solution is

van't Hoff factor,
$i = \frac{observed molar mass}{calculated molar mass}$
$= \frac{40}{25} = 1.6$
Since, sodium hydroxide is binary electrolyte,
$i = [1 + (2 - 1) α] (α =$ degree of ionisation)
$1.6 = (1 + α) α$
$= 1.6 - 1 = 0.6 = 60$