SATHEE: Solutions Question 387

Solutions Question 387

Question: The expression relating degree of dissociation of the weak electrolyte $A_{x} B_{y}$ with its van’t Hoff factor is

Options:

A) $α = \frac{i - 1}{x + y - 1}$

B) $α = \frac{i + 1}{x + y - 1}$

C) $α = \frac{x + y - 1}{i - 1}$

D) $α = \frac{x + y - 1}{i + 1}$

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Answer:

Correct Answer: A

Solution:

$\underset{1 - α}{A_{\times} B_{y}}, \underset{α x}{x A^{y +}}, + \underset{α y}{y B^{x -}},$ Total amount of species $= 1 + (x + y - 1) α$ Van’t Hoff factor, $i = \frac{1 + (x + y - 1) α}{1}$ or $α = \frac{i - 1}{x + y - 1}$

Solutions Question 387

Question: The expression relating degree of dissociation of the weak electrolyte $A_{x} B_{y}$ with its van’t Hoff factor is

Options:

Answer:

Solution:

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Solutions Question 387

Question: The expression relating degree of dissociation of the weak electrolyte AxBy with its van’t Hoff factor is

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The expression relating degree of dissociation of the weak electrolyte $A_{x} B_{y}$ with its van’t Hoff factor is