The ratio of molarity and normality of a dibasic acid solution is :

4

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2

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\frac{1}{5}

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\frac{1}{2}

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Solution

The correct option is C $\frac{1}{2}$
$N o r m a l i t y = m o l a r i t y \times e q u i v a l e n t H$

For dibasic acid solution, $e q u i v a l e n t H = 2$

$\frac{M o l a r i t y}{N o r m a l i t y} = \frac{1}{2}$

Hence, option D is correct.

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Similar questions

Q. For a dibasic acid, normality is twice the molarity for a given solution.

H_{2} X

is a dibasic acid which dissociates completely in water. Which one of the following is the molarity of an aqueous solution of this acid which has a

p H

1

Statement 1: 0.1 M $H_{3} P O_{3} (a q .)$ solution has normality equal to 0.3N when completely reacted with NaOH.
Statement 2 : $H_{3} P O_{3}$ is a dibasic acid.
Which of the above statement is correct?

Q. Regarding molarity, which of the following statements are correct?

a) Units of molarity gm-moles kg

^{- 1}

b) Molarity of dibasic acid is half of its normality

\frac{N o r m a l i t y \times G E W}{G M W}

d) Morality always equals to its molality

Q. If 20ml of 0.4 N NaOH solution completely neutralise 40ml of a dibasic acid, the molarity of the acid solution is:

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The ratio of molarity and normality of a dibasic acid solution is :

The correct option is C 12Normality=molarity×equivalent HFor dibasic acid solution, equivalent H=2MolarityNormality=12Hence, option D is correct.

The correct option is C $\frac{1}{2}$
$N o r m a l i t y = m o l a r i t y \times e q u i v a l e n t H$

For dibasic acid solution, $e q u i v a l e n t H = 2$

$\frac{M o l a r i t y}{N o r m a l i t y} = \frac{1}{2}$

Hence, option D is correct.