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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 .
From the N1V1 = constant concept we get,1*1/2 = 0.6*N2we get,N2 = 10/12 => 5/6 => 0.83This would be the final normality of the solution.
1L of N/2 HCL contains = ½ x 36.5 = 18.25g HCL now present = 18.5-3.25 = 15.25 which is approximately =15gNow, volume=600 ml= 0.600 LNow Normality= 15/36.5 x 1/0.600 = 0.68N Therefore the Normality is equal to 1.68 N
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 gAfter volume was reduced to 600 ml 3.25 g was lost, implies remaining HCl weight is 18.25 – 3.25 = 15gTherefore new normality = [(15/36.5)/600]x1000 = 0.6849 N
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 gHCl now present = 18.5- 3.25 g = 15.25 gapproximately, = 15 gnow volume = 600 ml = 0.006Lnow normality= 15/36.5× 1/ 0.006 g= 0.68 gtherefore, the normality is equal to 1.68 N ( ans)Thanks and Regards
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