Making 1 N solution of NaOH To make 1 N solution, dissolve 40.00 g of sodium hydroxide in water to make volume 1 liter. For a 0.1 N solution (used for wine analysis) 4.00 g of NaOH per liter is needed.

1 Expert Answer You would add 13.6 g NaCl to 250 mL of water.

Answer is: dissolve 1.46g of NaCl in 50mL H2O and dilute to 250mL.

To prepare a 5 M solution: Dissolve 292 g of NaCl in 800 mL of H2O. Adjust the volume to 1 L with H2O. Dispense into aliquots and sterilize by autoclaving. Store the NaCl solution at room temperature.

A 2.5 M solution is 2.5 moles per liter (Molarity is just the number of moles per liter). Therefore, 0.5 L would contain 1.25 mol. Hence, you would need 1.25 × 58.44 g = 73 g.

A 1M solution would consist of 342.3 grams sucrose in one liter final volume. A concentration of 70 mM is the same as 0.07 moles per liter. Take 0.07 moles/liter times 342.3 grams per mole and you have 23.96 grams needed per liter. To make 200 milliliters of your solution multiply grams/liter by liters needed.

Terms in this set (7)

1. Determine the mass of the solute needed to make the solution (calculate it)
2. Select the correct size volumetric flask for the quantity of solution.
3. – Measure out the solute in a small beaker.
4. – Pour the solution into the volumetric flask.

Use the following steps for diluting a stock solution:

1. Use a volumetric flask.
2. Add some of your solvent to the flask, but not all of it.
3. Measure out the needed amount of stock solution.
4. Add your measured stock solution to the volumetric flask.
5. Carefully fill the flask to the marked line with solvent.

Mix with powdered compound until dissolved.

1. For example: Mix 500 mL of water and 25 g of NaCl to make a 5% solution.
2. Remember, if you’re diluting a liquid compound, you must subtract out the volume of liquid being added from the final volume: 500 mL – 25 mL = 475 mL of water.

Hydrochloric acid Solution Preparation Take about 100 ml of water in a cleaned and dried 1000 ml volumetric flask. Add about 8.5 ml of Conc. Hydrochloric acid with continuous stirring. Add more about 700 ml of water, mix and allow to cool to room temperature.

i.e You need to take 0.821 ml of reagent and make it to 100ml with de ionized water in order to get 0.1N 100ml HCl standard solution. The prepared HCl solution is an approximate solution of 0.1N. To standardise it should be titrated against Sodium bicarbonate standard solution.

The GMW of HCl would be the atomic weight of H added to the atomic weight of Cl: H = 1 + Cl = 35.45 = 36.45 g. A liter of 1M solution of HCl would contain 36.45 g.

1.3 * 106

To make a 10 mM 2,4,6-tri(2-pyridyl)-1,3,5-triazine (TPTZ) solution, add 0.31 g to 100 mL HCl (40 mM). This solution is stable for at least 1 month at 4°C.

Step 3: Calculate the concentration of hydrochloric acid in mol/dm 3

1. Concentration in mol/dm 3 =
2. Concentration in mol/dm 3 =
3. = 0.125 mol/dm 3
4. Relative formula mass of HCl = 1 + 35.5 = 36.5.
5. Mass = relative formula mass × amount.
6. Mass of HCl = 36.5 × 0.125.
7. = 4.56 g.
8. So concentration = 4.56 g/dm 3