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?

To prepare a 0.5N solution of HCl using a 37% HCl solution and a 0.1N HCl solution, you need to perform a dilution calculation.

Here's how you can do it:

1. Determine the amount of 37% HCl solution needed:
- The 37% HCl solution is not in N (Normality), so you need to convert its concentration to N. Since the density of the 37% HCl solution is not provided, we will assume it to be 1.2 g/mL.
- To convert the 37% HCl concentration to N, you need to calculate the molarity (M). Remember, Molarity (M) is the number of moles of solute per liter of solution.
- The molar mass of HCl is 36.46 g/mol.
- First, calculate the number of moles of HCl in 1000 mL (1 liter) of the 37% HCl solution:
Moles of HCl = (Percentage of HCl / 100) * (Density of Solution / Molar Mass of HCl)
= (37 / 100) * (1.2 g/mL / 36.46 g/mol) * (1000 mL)
= 12.118 moles of HCl
- Now, calculate the number of moles of HCl needed for a 0.5N solution:
Moles of HCl needed = 0.5 N * (1 liter) = 0.5 moles of HCl
- Divide the moles of HCl needed by the moles of HCl in the 37% solution to determine the volume needed:
Volume of 37% HCl solution needed = Moles of HCl needed / Moles of HCl in 37% solution
= 0.5 moles of HCl / 12.118 moles of HCl
= 0.0412 liters or 41.2 mL

2. Determine the amount of 0.1N HCl solution needed:
- Since the 0.1N HCl solution is already in N, you only need to figure out the volume needed to achieve 0.5N.
- The formula used here is M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
- Rearrange the formula to solve for V2, the volume needed:
V2 = (M1V1) / M2
V2 = (0.1 N * V1) / 0.5 N
V2 = (0.1 * V1) / 0.5
V2 = 0.2 * V1

Therefore, the volume of the 0.1N HCl solution needed to get a 0.5N solution is 0.2 times the required final volume.

3. Final steps:
- Mix the calculated volume of the 37% HCl solution (41.2 mL) with the calculated volume of the 0.1N HCl solution (0.2 * final volume).
- Adjust the total volume to the desired final volume by adding appropriate amounts of solvent, which is usually water.
- Mix thoroughly to ensure homogeneity.

Following these steps will help you prepare a 0.5N HCl solution using a 37% HCl solution and a 0.1N HCl solution.

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.

How do I prepare 0.5 molar ethanolic Potassium hydroxide

-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