How to make 1M concentrated HCl
Ah - chemistry class 2004/2005 11ch1
Example: acetic acid in different concentrations - as vinegar essence (25%) or table vinegar (5%) or acid in a dropper bottle (2 mol / l)
Conversion of the concentration data mol / l <-->%
The percentage is given in g / ml as a sufficient approximation.
For acetic acid, 2 mol / l means that one liter of the solution contains 2 mol of acetic acid, i.e. about twice 6.022 * 1023 Acetic acid molecules. 2 mol acetic acid (CH3COOH) have a mass of 120 g (molar mass of acetic acid: 2 * 12 g / mol (C) + 4 * 1 g / mol (H) + 2 * 16 g / mol (O) = 60 g / mol)
|Amount of substance = existing substance mass / molar mass||n = m / M [mol]|
|Concentration = amount of substance / volume||c = n / V [mol / l]|
For acetic acid: 2 mol / l = 120 g / l = 12 g / 100 ml,
2 mol / l therefore corresponds to an acetic acid concentration of approx. 12%.
Determination of the molar concentration of vinegar essence and table vinegar
If the specification of 2 mol / l for acetic acid corresponds to a concentration of 12%, the table vinegar would have to be a little less than 1 molar and the vinegar essence a little more than 4 molar.
Table vinegar: 5% corresponds to 5 g / 100 ml = 50 g / l. 50 g of acetic acid are 0.83 mol of acetic acid, so the table vinegar has a concentration of 0.83 mol / l.
Vinegar essence: 25% corresponds to 25 g / 100 ml = 250 g / l. 250 g of acetic acid are 4.17 mol of acetic acid, so the vinegar essence has a concentration of 4.15 mol / l.
- Prepare sodium hydroxide solution with a concentration of 2 mol / l
- M (NaOH) = 40 g / mol, so 2 mol NaOH are 80 g, which is then with water filled up to one liter Need to become.
- Making silver nitrate from silver and nitric acid
- 2 (mol) Ag + 2 (mol) HNO3 -> 2 (mol) AgNO3 + (1 mol) H2
- Prepare a 0.1 molar silver nitrate solution
- M (AgNO3) = 108 g / mol + 14 g / mol + 3 * 16 g / mol = 170 g / mol
- n = 0.1 mol so m = n * M = 0.1 mol * 170 g / mol = 17 g
- So 17 g of silver nitrate must be mixed with water made up to 1 l of solutionbecome.
- Prepare 1 molar hydrochloric acid from 37% hydrochloric acid
- Hydrochloric acid: HCl (aq) <-> H3O+ + Cl-
- M (HCl) = 36 g / mol 37% = 37 g / 100 ml = 370 g / l
n = m / M = 370 g / 36 g / mol = 10.3 mol
- To a good approximation, one can assume that 37% hydrochloric acid is 10 molar.
- If 1 liter of 37% hydrochloric acid contains 10 mol of HCl, 100 ml contains 1 mol of hydrochloric acid. So for a 1 molar hydrochloric acid, 100 ml of 37% made up to 1 liter become.
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