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Standard XII

Chemistry

Abnormal Colligative Properties

1L solution o...

Question

$1 L$ solution of a diacidic base was prepared by dissolving $5.30 g$ of it in water. For complete neutralization $30 m L$ of this solution required $18 m L$ of $N / 6$ $H C l$ . Calculate equivalent mass and molecular mass of the base.

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Solution

For neutralisation reaction
(complete)
eq. of acid $=$ eq. of base
$∴$ conc of diacidic base $\times$ volume $=$ conc of acid $\times$ volume of acid
$x \times 30 m l = \frac{N}{6} \times 18$ ml
$x = \frac{N}{10}$ (conc of diacidic base)
Since it's diacidic base
$∴$ Its Molarity $= \frac{M}{20}$ (Acidity $= 2$ )
[Using: Normality $=$ molarity $\times$ acidity]
It's molarity is $\frac{1}{20}$ , that means $1$ litre solution contains $\frac{1}{20}$ moles of the base.
And according to the question, $\frac{1}{20}$ moles have weight $5.30$ g.
$∴$ $1$ mole $\to 20 \times 5.3$
$= 106$ g $=$ molar mass
$∴$ Eq mass $= \frac{m o l a r m a s s}{a c i d i t y} = \frac{106}{2} = 53$ g.

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1L solution of a diacidic base was prepared by dissolving 5.30g of it in water. For complete neutralization 30mL of this solution required 18mL of N/6 HCl. Calculate equivalent mass and molecular mass of the base.

$1 L$ solution of a diacidic base was prepared by dissolving $5.30 g$ of it in water. For complete neutralization $30 m L$ of this solution required $18 m L$ of $N / 6$ $H C l$ . Calculate equivalent mass and molecular mass of the base.