How to make a 0.1 normal solution?
Answer:
This is very easy------
since equivalent mass of acid= molar mass/basicity
Now molar mass of CH3COOH= 60 g/mol
and its basicity is 1
So u get equivalent mass = 60/1 = 60 g
This means that 60 g of the acid is dissolved in one litre of the water.
But in the question u r asked to prepare 0.1 normal solution.
So 1 normal = 60 g/litre
0.1 normal = 60/10
= 6 g/ litre
(hint= 0.1N means 1N/10)
1 eq of CH3COOH = 1 mol of CH3COOH (since its basicity =1)
= 60 gm CH3COOH.
=> 0.1 eq of CH3COOH = 6 gm CH3COOH
So, u need to take 6 gm of glacial acetic acid in a graduated conical flask, & pour water in it till the level reaches 1 lt mark . Shake well to dissolve the acid completely in water. U'll have 1 lt of 0.1 N acetic acid when the acid completely get dissolved in water.
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