SATHEE: Equilibrium Question 830

Equilibrium Question 830

Question: If degree of dissociation of pure water at $100^{\circ} C$ is $1.8 \times 10^{- 8}$ , then the dissociation constant of water will be (density of $H_{2} O = 1 g / c c$ )

Options:

A) $1 \times 10^{- 12}$

B) $1 \times 10^{- 14}$

C) $1.8 \times 10^{- 12}$

D) $1.8 \times 10^{- 14}$

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Answer:

Correct Answer: D

Solution:

As, molarity, $= \frac{W t . o f s o l u t e p e r l i t r e o f s o l u t i o n}{M o l . w t . o f s o l u t e}$ Molarity of $H_{2} O = \frac{1000}{18} m o l / L$

$\underset{c (1 - α)}{H_{2} O}, \to \underset{c α}{H^{+}}, + \underset{c α}{O H^{-}},$ Thus, $K_{a} = \frac{c α^{2}}{1 - α} = c α^{2}$

$= \frac{1000}{18} \times {(1.8 \times 10^{- 8})}^{2}$

$= 1.8 \times 10^{- 14}$

Equilibrium Question 830

Question: If degree of dissociation of pure water at $100^{\circ} C$ is $1.8 \times 10^{- 8}$ , then the dissociation constant of water will be (density of $H_{2} O = 1 g / c c$ )

Options:

Answer:

Solution:

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Equilibrium Question 830

Question: If degree of dissociation of pure water at 100∘C is 1.8×10−8 , then the dissociation constant of water will be (density of H2O=1g/cc )

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: If degree of dissociation of pure water at $100^{\circ} C$ is $1.8 \times 10^{- 8}$ , then the dissociation constant of water will be (density of $H_{2} O = 1 g / c c$ )