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1. Weigh 58,4 mg 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.
What else can I help you with?
How much sodium chloride is required to make 1litre of 1M solution?
To make a 1M solution of sodium chloride in 1 liter of water, you would need 58.44 grams of sodium chloride. This is based on the molecular weight of sodium chloride, which is 58.44 g/mol.
How many moles of sodium ions and chloride ions are present in 1 liter of a 1M solution of sodium chloride?
In a 1M solution of sodium chloride, there would be 1 mole of sodium ions and 1 mole of chloride ions in 1 liter of the solution. This is because each formula unit of sodium chloride dissociates into one sodium ion and one chloride ion in solution.
Which phase will be on top in an immiscible mixture of methylene chloride and 1M of sodium hydroxide?
Sodium hydroxide solution will be on the top.
How can we prepare 1M solution of NaCl?
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
What does 1M solution of sodium carbonate mean?
A 1M solution of sodium carbonate means that it contains 1 mole of sodium carbonate dissolved in 1 liter of solvent (usually water). This concentration is used in chemistry to describe the amount of the solute (sodium carbonate) present in the solution.
What is produced at the cathode of an electrolytic cell containing 1M solution of sodium chloride?
Chlorine gas. Cl2
Ionic formula for sodium chloride?
The ions in sodium chloride are Na+1 and Cl-1.
How to prepare 1m hcl from 35 percent hcl?
To prepare 1M HCl solution from 35% HCl solution, you would need to dilute the 35% HCl with water. Use the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the volume of the initial solution, C2 is the final concentration (1M), and V2 is the final volume (1 liter in this case). Calculate the volume of 35% HCl needed to achieve a 1M solution, then add water to make up the total volume to 1 liter.
How do you prepare 1M sodium phosphate dibasic?
sodium phosphate dibasic means that there are two sodium molecules in the formula (contributing to two bases). the formula is Na2HPO4. the first step is to figure out the molecular weight of the compount with the help of a periodic table. This comes out to 142g/mole. To make 1M solution (M= moles/liter so I am assuming 1 liter) you need 142 grams of the compound in 1L of dH2o. The definition of molarity (which is moles over liters) makes the calculation easy for making 1L of the solution. However if you want to make less or more of the solution, you will have to calculate accordingly. For example if you wanted to make 50 ml (0.05L) of the solution, you would multiply 142 g/liter (g/mol x moles/liter will give g/liter) X 0.05 liter, which would give you 7.1 g. 7.1 g of Na2HPO4 is needed to make 50 ml of 1 M solution. 142 g is needed to make 1000ml (1 L) of 1M solution.
How would you make a 100ml of a Mole solution of Sodium Chloride?
To make a 100ml 1M solution of Sodium Chloride, you would dissolve 5.85 grams of NaCl in enough water to make 100ml of solution. This molarity calculation is based on the molar mass of NaCl (58.44 g/mol).
How do you prepare a solution containing 300ml of 1M Sodium Sulfate?
That is a poorly worded question. The way you have worded it could be answered correctly by instructing you to "Pour 300ml of 1M solution in a swimming pool."The question is usually posed in the form of "How much sodium sulfate is needed to produce 300 ml of 1M solution?" Presuming you intend the latter, you would prepare 300 ml of 1M solution by mixing 300/1000 of a mole of sodium sulfate with sufficient water to produce 300 ml of solution.Note the difference between Molar and Molal: 1M = 1 mole per liter of solution, 1 molal = 1 mole per liter of water. This distinction has a major effect on the wording of your answer.300 ml / 1000 ml = 3/10 = 0.3000One mole of sodium sulfate (Na2SO4) has the formula weight of 142.04314 g/mol as shown below:fw Na x 2 = 22.98977 g/mol x 2 = 45.97954 g/molfw S = 32.066 g/molfw O x 4 = 15.994 g/mol x 4 = 63.9976 g/mol45.97954 g/mol + 32.066 g/mol + 63.9976 g/mol = 142.04314 g/molThe mass of 0.3000 mol x 142.04314 g/mol = 42.613 g.Note: The final answer was rounded to the thousandths place because the formula weight for Sulfur was the least precise term.
What is the make up of 0.1 sodium hydroxide?
0.1 M sodium hydroxide solution contains 0.1 moles of sodium hydroxide per liter of solution. This corresponds to 0.1 moles of NaOH per 40 g (1 mole) of NaOH, resulting in 4 g of NaOH in 1 liter of 0.1 M NaOH solution.
