Question

10ml of hcl solution gave 0.1435gm of agcl when treated with excess of agno3 the normality of the hcl solution is

Answer 1

HCl + AgNO3 …............>AgCl+ HNO3
10ml …...excess .... ….............0
0ml …..excess.. …..................0.1435g
...................................................=(0.1435/143.5) mol
=0.001mol AgCl made by 10 ml of HCl
From reaction it is clear that,
1 mole of AgCl would be obtained by1 mole HCl
So number of moles of HCl in 10ml would be=0.001
SO number of equivalents of HCl in 10ml = 0.001*1 …..(as basicity of HCl =1)
Hence normality of HCl=(number of equivalents of HCl/volume of HCl in ml)*1000
normality= (0.001/10)*1000=0.1N