Question

6.
One litre of N/2 HCl solution is heated in a beaker. It was
observed that when the volume of the solution was reduced to
600 ml, 3.25g of HCI is lost. Calculate the normally of the new
solution​

Answer 1

Answer:

According to the question,

1litre of N/2 HCl contents = 1/2× 36.5

= 18.26 g

HCl now present = 18.5- 3.25 g = 15.25 g

approximately, = 15 g

now volume = 600 ml = 0.006L

now normality= 15/36.5× 1/ 0.006 g

= 0.68 g

therefore, the normality is equal to 1.68 N ( ans)

Answer 2

Answer:

According to the question ,

N/2 HCl = 1/2 X (35.5 + 1 ) =>18.25gm

Now,

3.25gm of HCl is given out.

So,

remaining amount = 18.25-3.25

= 15 gm

This the weight of HCl solution .

GMW = 36.5 gm

Volume = 600 ml => 0.6 L

Now,

Normality= wt/GMW + 1/V

=> 15/36.5 + 10/6

=> 2.0776 gm

Explanation: