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 .

Question

Arun · Answer

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.

Vudhanthi Neeraja · Answer

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

Vishal P · Answer

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

Rishi Sharma · Answer

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)

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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 .

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 .

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