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

Equivalent Mass

Calculate the...

Question

Calculate the equivalent weight of $K H C_{2} O_{4}$ (potassium acid oxalate) when it is neutralised by $K O H$ . Give answer in form of $\frac{E q . w t}{64} = x$ i.e. find the value of $x$ .

Open in App

Solution

The neutralization reaction is as given below.
$K H C_{2} O_{4} + K O H ⟶ K_{2} C_{2} O_{4} + H_{2} O$
$K H C_{2} O_{4} \equiv 1 H^{+} \equiv 1 {O H}^{-} (H^{+} i o n i s n e u t r a l i z e d)$
Thus, the equivalent weight will be equal to its molecular weight.
$E q u i v a l e n t w e i g h t = \frac{M o l e c u l a r w e i g h t}{1} = \frac{128}{1} = 128$

Now,

$\frac{E q . w t .}{64} = \frac{128}{64} = 2 = x$ .

Suggest Corrections

Similar questions

K H C_{2} O_{4}

100

0.02 M K M n O_{4}

M n^{2 +})

x

100

0.05 M C a (O H)_{2}

y

4.2 g

K O H

C a (O H)_{2}

0.1 g r a m e q u i v a l e n t s

K_{2} C_{2} O_{4} .3 H_{2} C_{2} O_{4} .4 H_{2} O

M n O_{4}^{-}

0.3 M M n O_{4}^{-}

10.16 g

508 g m o l^{- 1}

K_{2} C_{2} O_{4} . 3 H_{2} C_{2} O_{4} .4 H_{2} O

M n O_{4}^{-}

0.1 M K M n O_{4}

5.08 g m

63 g/eq.

\frac{N}{10}

Join BYJU'S Learning Program