how can we prepare 0.5 n solution of hcl as we have 37% hcl and 0.1n hcl. and we want to make 0.5n hcl from 0.1n hcl by pouring 37% hcl?

Answer 1

To begin, I don't think this is possible using standard volumetric equipment. I'll explain below.
Let's first determine the N of the 37% stuff.
The density of 37% HCl is 1.19g/mL so the N is
1000 mL x 1.19 g/mL x 0.37 x (1/36.5) = about 12N for the 37% HCl.
How much 0.5N HCl do you wish to prepare. Let's say 100 mL so milliequivalents(me) HCl in 100 mL of 0.5N = 50 me. We have how many me in the 100 mL of 0.1N? That's 100 x 0.1 = 10. That leaves 40 me to add. N = me/mL or mL = me/N = 40/12 = 3.33 mL of the 12N. So let's see what the concn is.
100 mL x 0.1N = 10 me + 3.33 mL x 12N = 40 me = 50 me in a total volume of 103.33 = 50/103.33 = about 0.48N. If you don't want EXACTLY 0.5N this will do it.
There is another way to get exactly 0.5N.
Say we want to prepare 200 mL 0.5N which is 100 me. We start by added 100 mL x 0.1N = 10 me HCl. That makes us 90 ne short of our goal of 100 so we add 90/12 = 7.5 mL 12N HCl to that (7.5 x 12N = 90 me). That gives us our 100 me and we have all of that in a 200 mL volumetric flask. The volume is now about 100 + 7.5 = 107.5. We add water to the 200 mL mark to make EXACTLY 0.5N BUT we had to add water which is not allowed if I follow the problem correctly. Again, I don't think you can start with 0.1N HCl, pour in 37% HCl and produce EXACTLY 0.5N. You might be able to do it with grams but(in which you start with so many grams of the 0.1N and add dropwise the 37% HCl until a certain weight is achieved) but I didn't investigate that possibility.

Answer 2

How do I prepare 0.5 molar ethanolic Potassium hydroxide

Answer 3

-potassium hydroxide (pellets),KOH molecular weight - 56.1 g
-hence 0.5 M, 28 g(1/2 of 56) of KOH dissolve in 280 ml ethanol,
*take ratio 1 g:10 ml the it will be 28 g KOH: 280 ml ethanol
-stir them up till clear