20g of a binary electrolyte (mol. wt =100) are dissolved in 500 g of water. The depression in freezing point of the solution is 0.74oC ( Kf=1.86 text K molality-1 ) The degree of ionization of the electrolyte is:

Question

20g of a binary electrolyte (mol. wt =100) are dissolved in 500 g of water. The depression in freezing point of the solution is 0.74oC ( Kf=1.86 text K molality-1 ) The degree of ionization of the electrolyte is: (A)  0%  (B)  100%  (C)  75%  (D)  50%

Tardigrade Admin · Accepted Answer

Δ T=(1000× 1.86× 20/m× 500) 0.74=(1000× 1.86× 20/m× 500) m=100 Actual molecular mass = 100 ∴ The degree of ionization of the electrolyte is 0%.

Q. 20g of a binary electrolyte (mol. wt =100) are dissolved in 500 g of water. The depression in freezing point of the solution is $0.74^{o} C$ ( $K_{f} = 1.86 K$ $m o l a l i t y^{- 1}$ ) The degree of ionization of the electrolyte is:

$0$

$100$

$75$

$50$

$Δ T = \frac{1000 \times 1.86 \times 20}{m \times 500}$ $0.74 = \frac{1000 \times 1.86 \times 20}{m \times 500}$ $m = 100$ Actual molecular mass = 100 $∴$ The degree of ionization of the electrolyte is 0%.

Q. 20g of a binary electrolyte (mol. wt =100) are dissolved in 500 g of water. The depression in freezing point of the solution is 0.74oC ( Kf​=1.86K molality−1 ) The degree of ionization of the electrolyte is:

0

100

75

50

ΔT=m×5001000×1.86×20​ 0.74=m×5001000×1.86×20​ m=100 Actual molecular mass = 100 ∴ The degree of ionization of the electrolyte is 0%.

Q. 20g of a binary electrolyte (mol. wt =100) are dissolved in 500 g of water. The depression in freezing point of the solution is $0.74^{o} C$ ( $K_{f} = 1.86 K$ $m o l a l i t y^{- 1}$ ) The degree of ionization of the electrolyte is:

$0$

$100$

$75$

$50$

$Δ T = \frac{1000 \times 1.86 \times 20}{m \times 500}$ $0.74 = \frac{1000 \times 1.86 \times 20}{m \times 500}$ $m = 100$ Actual molecular mass = 100 $∴$ The degree of ionization of the electrolyte is 0%.