How do you prepare a solution of 6N acetic acid from glacial acetic acid?

Answer 1

Molecular Weight of Acetic acid=60.05

density: 1.049 gr/cm3

0.5 molar=60.05/2= 30.025 gr

you have molecular weight and density of Acetic acid so you can find the volume:

d=m/v => v=30.025/1.049= 28.62 ml

Add 28.62 ml of acetic acid in a beaker and complete to 1L by adding 971.38 ml water, you will have 1 L of 0.5 M acetic acid solution.

Answer 2

Since glacial acetic acid is 17.4N, you just use (M1)(V1) = (M2)(V2).

Example: If you want to prepare 100mL of 6N: (17.4N)(x mL) = (6N)(100 mL). So, you need 34.5 mL of glacial, dilute to 100 mL with water.

Answer 3

Concentration of glacial acetic is 99.8%

Density is 1.049 g/ml

Molecular mass is 60.05 g/mol

1000 mlglacial acetic contains 1049x0.998=1046.902 g

It can calculate in term of Molarity =1046.902 (g)/(60.05 (g/mol))*1 liter =17.43 M

For acetic acid 1 M = 1 N,Normality of Glacial acetic (99.8 %) = 17.43 N

C1V1=C2V2

17.43(N)*V1=0.1*1000ml

V1= 5.73 ml

So, 5.73 mlglacial acetic is added into water adjusted to 1 L to obtain 0.1 N

Answer 4

Its simple, use this formula V1N1=V2N2,

where N1=17.4N (Concentration of acetic acid)

N2= 0.5 (/the concentration that we need)

V2= 1000mL (for a litre of solution)

substitute this in the equation, as below

0.5*1000/17.4

then we get 28.7,

So you need to add 28.7 mL of acetic acid to 1000 mL of DI water to make 0.5 N acetic acid soution.

Answer 5

Glacial acetic acid is concentrated acetic acid. Therefore we can just dilute it with water

Answer 6

If you are wanting to make 0.1 M acetic acid you will need 576 microliters of glacial acetic acid in a 10 ml solution with water.

Answer 7

use 2 units of acetic acid for example 2ml in 98 units of water

Answer 8

Dissolve 0.2 mol of CH3COOH powder i.e. 12 g in 1 L water.

Answer 9

It is impossible to transform the acetic acid in nitric acid.