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Grade: 12th pass

                        

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

2 years ago

Answers : (4)

Arun
24742 Points
							
 
From the N1V1 = constant concept we get,
1*1/2 = 0.6*N2
we get,
N2 = 10/12 => 5/6 => 0.83
This would be the final normality of the solution.
2 years ago
Vudhanthi Neeraja
13 Points
							
1L of N/2 HCL contains = ½ x 36.5 = 18.25g 
HCL now present = 18.5-3.25 = 15.25 which is approximately =15g
Now, volume=600 ml= 0.600 L
Now Normality= 15/36.5 x 1/0.600 = 0.68N 
Therefore the Normality is equal to 1.68 N
one year ago
Vishal P
24 Points
							
N/2 HCl means ½ N HCL, i.e, 0.5 N HCl.
Implies is one litre of solution you have 0.5 moles of HCl, i.e, 36.5/2 g of HCl=18.25 g
After volume was reduced to 600 ml 3.25 g was lost, implies remaining HCl weight is 18.25 – 3.25 = 15g
Therefore new normality = [(15/36.5)/600]x1000 = 0.6849 N
11 months ago
Rishi Sharma
askIITians Faculty
614 Points
							Dear Student,
Please find below the solution to your problem.

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)

Thanks and Regards
2 months ago
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