Question

calculate the molarity of 6.3g of oxalic acid (H2C2O4 2H2O)present in 500 ml of solution. (molecular weight of oxalic acid is 126)​

Answer 1

Answer:

Normality is defined as the number of mole equivalents per litre of solution.

Oxalic acid C

2

H

2

O

4

.2H

2

O

Molecular mass =126g

Equivalent weight =

2

126

=63g/equiv

Normality =

Eq. wt. ×1000

weight

=

63×

1000

500

6.3

=0.1×2=0.2N

Answer 2

Question:

• Calculate the molarity of 6.3g of oxalic acid (H2C2O4 2H2O)present in 500 ml of solution. (molecular weight of oxalic acid is 126).

Solution:

As we know that,

• M = $\Large \rm{\frac{w}{GMW} \times \frac{1000}{V_{(ml)}} }$

Given:

• w = 6.3g
• GMW = 126g
• $\rm{V_{(ml)} = 500ml}$
• M = ?

Applying the values,

➽ $\because \rm \: {M = \Large\cancel\frac{6.3}{126} \times\cancel\frac{1000}{500} }$

➽$\rm{M = 0.1}$

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