You visited us 1 times! Enjoying our articles? Unlock Full Access!

Abnormal Colligative Properties

1 molal aqueo...

Question

$1 m o l a l$ aqueous solution of an electrolyte $A_{2} B_{3}$ is $60 %$ ionised. The boiling point of the solution at $1 a t m$ is $K$ . (Rounded-off to the nearest integer) [Given: $K_{b}$ for $(H_{2} O) = 0.52 K k g m o l^{- 1}]$

375.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

375

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

375.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

375

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

374.76

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

374.77

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

$A_{2} B_{3} \to 2 A^{3 +} + 3 B^{2 -}$
No. of ions $= 2 + 3 = 5$
$i = 1 + (n - 1) α i = 1 + (5 - 1) \times 0.6 i = 1 + 4 \times 0.6 = 1 + 2.4 = 3.4$

$Δ T_{b} = K_{b} \times m \times i Δ T_{b} = 0.52 \times 1 \times 3.4 = {1.768}^{\circ} C$
$Δ T_{b} = (T_{b})_{s o l u t i o n} - (T_{b})_{H_{2} O} 1.768 = (T_{b})_{s o l u t i o n} - 100 (T_{b})_{s o l u t i o n} = {101.768}^{\circ} C \approx 375 K$

Suggest Corrections

Similar questions

1 m o l a l

A_{2} B_{3}

60 %

1 a t m

K

K_{b}

(H_{2} O) = 0.52 K k g m o l^{- 1}]

A_{2} B_{3}

60 %

[K_{b (H_{2} O)} = 0.52 K k g m o l^{- 1}]

1.0 molal

A B_{4}

75 %

(in K)

1 atm

K_{b (H_{2} O)} = 0.5 K kg {mol}^{- 1}

$1 molal$ aqueous solution of an electrolyte $A_{2} B_{3}$ is $60 %$ ionised. The boiling point of the solution at $1 atm$ is $_____K$ .

0.1 molal

A B_{4}

75 %

in K

1 atm

K_{b (H_{2} O)} = 0.52 K kg {mol}^{- 1}

Join BYJU'S Learning Program