Question

How much oxalic acid is required to prepare 100ml 0.2n?

Answer 1

since normality is equal to. 0.2 and molarity is equal to 0.2/2 which is equal to 0.1 therefore oxalic acid required to prepare 100 ml point 2 normal solution is equal to 1 litre

Answer 2

Answer:

The weight of oxalic acid required to prepare the $100ml$ solution = $1.26g$.

Explanation:

The normality of the oxalic acid solution, N = $0.2N$

The volume of the oxalic acid solution, V = $100ml$

The weight of oxalic acid required to prepare the $100ml$ solution =?

From the given relation, we can find out the weight of oxalic acid:

• Normality = $\frac{weight\times 1000}{Equivalent weight \times volume (L)}$

For this, we have to find out the equivalent weight of oxalic acid:

As we know, oxalic acid usually occurs as the dihydrate ( $H_{2}C_{2}O_{4}.2H_{2}O$ )

• Equivalent weight = $\frac{Molar weight}{n-factor}$

For oxalic acid:

• The molecular weight of $H_{2}C_{2}O_{4}.2H_{2}O$ = $126g/mol$

And n-factor is the number of $H^{+}/OH^{-}$ ions replaced by 1 mole of acid/base in a reaction.

• n-factor for $(COOH)_{2}$ = $2$ ( as 2 $H^{+}$ ions can be replaced )

Therefore,

• Equivalent weight = $\frac{Molar weight}{n-factor}$ = $\frac{126}{2}$ = $63g/mol$

Now,

• Normality = $\frac{weight\times 1000}{Equivalent weight \times volume (ml)}$
• $0.2 = \frac{weight \times 1000}{63\times 100}$
• $1000\times weight = 1260$
• weight = $1.26g$

Hence, the weight of oxalic acid required to prepare the $100ml$ solution = $1.26g$.