Question

12.5 ml of a solution containing 6 grams of dibasic acid in 1 litre was found to be neutralized by 10 ml of a decinormal solution of NAOH .the molecular mass of the acid is

Answer 1

Answer: Amount of acid in 1L (1000 ml) solution = 6 g

Amount of acid in 12.5 ml solution = (6/1000) x 12.5 g

 = 0.075 g

At neutralisation,

10 ml of N/10 NaOH ≡≡10 ml N/10 acid

Now,

10 ml N/10 acid = 0.075 g

So, 1000 ml N/10 acid = (0.075/10) x 1000

 = 7.5 g

 1000 ml of N acid = 7.5 x 10 = 75 g

Thus the equivalent mass of acid = 75

So, molecular mass = 75 x 2 = 150 g  ( since the acid is dibasic).

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Answer 2

Answer:

DIBASIC ACID:

6g-----------> 1000 ml

?g-------------> 12.5 ml

=> 0.075g

Use NORMALITY formula:

N= weight in grams /Gram equivalent weight × 1000/Volume in ml

Gram equivalent weight = weight in grams/ Normality ×1000/Volume in ml

THEREFORE,

Gram equivalent weight = 0.075/0.1 ×1000/10

= 75 gram

Gram equivalent weight = gram molecular weight/ basicity

gram molecular weight = 75×2

= 150 gm

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