22. Calculate the amount of Cu+2 in a 150 ml blue vitriol solution if 10 ml of it is titrated with 5 ml 0.039N sodium thiosulphate solution. (1 ml 1N Na2S2O3  0.06354 gm of Cu+2). Ans: 0.18585 gm

To calculate the amount of Cu+2 in a 150 ml blue vitriol solution, we need to follow these steps:

1. First, we need to understand the relationship between the sodium thiosulphate solution and Cu+2. According to the problem, 1 ml of 1N Na2S2O3 is equivalent to 0.06354 gm of Cu+2.

2. Next, we need to calculate the amount of Cu+2 in the 10 ml of blue vitriol solution that was titrated. We do this by multiplying the volume of the sodium thiosulphate solution used (5 ml) by its normality (0.039N) and by the equivalent weight of Cu+2 (0.06354 gm/ml).

   So, Cu+2 in 10 ml solution = 5 ml * 0.039N * 0.06354 gm/ml = 0.01239 gm.

3. Now, we need to find out how much Cu+2 would be in the entire 150 ml of blue vitriol solution. We do this by setting up a proportion. If 10 ml contains 0.01239 gm of Cu+2, then 150 ml would contain how much?

   So, Cu+2 in 150 ml solution = (0.01239 gm * 150 ml) / 10 ml = 0.18585 gm.

Therefore, the amount of Cu+2 in a 150 ml blue vitriol solution is 0.18585 gm.