1. Weigh 5,845 g ultrapure NaCl dried at 110 0C for 30 min.
2. Transfer NaCl in a clean 1 L volumetric flask using a funnel. 3. Wash the funnel with 0,9 L demineralized water.
4. Put the flask in a thermostat and maintain 30 min at 20 0C.
5. Add demineralized water up to the mark.
6. Stir vigorously and transfer in a clean bottle with stopper.
7. Add a label with necessary information.
Dissolve 100 mg NaCl in 1.0 Litre water.
Dissolve 2g of NaCl in 100 cm3 water at normal temperature.
To prepare a 0.9% solution take 0.9grams NaCl and dilute with 100mls of water.
The volume is 342 mL.
By a slowly evaporation of water from a NaCl solution.
You have to evaporate (by open boiling) 45 mL of the 75 mL 2M NaCl solution thus reducing the volume to 30 mL 5M NaCl.
Dissolve 16,61 g NaCl analytical reagent, dried in 1 L of demineralized water at 20 0C, in a volumetric flask.
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
Sodium chloride is the result of this reaction: NaOH + HCl = NaCl + H2O
Reactions are: 2Na + Cl2 = 2NaCl NaOH + HCl = NaCl + H2O
A method is the following:NH4Cl + NaNO2 = N2 + NaCl + 2 H2O
We first calculate the amount, in moles, of NaCl that we will need.Amount of NaCl needed = 0.24 x 400/100 = 0.096mol. Mass of NaCl needed = (23.0 + 35.5) x 0.096 = 5.616g So to produce 400ml of 0.24M NaCl solution, accurately add 5.616 grams of NaCl to 400ml of deionised water.