1.Weigh 292,2 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.
One mole of NACl (also known as table salt) is 58.5 gram so you need 0.0585 gram of NaCl in one litter.
The answer is 20 microliters.
If you need to make just 100mL, then you need 1 tenth of a liter that is 5M. If you were to make 1L of 5 molar NaCl, you would need 5 times the molar mass of NaCl (58.44g/mol) dissolved in 1L of water. Thus for 1L of a 5M solution you need 5 * 58.44g, or 292.2 grams of NaCl. However, since we only want 100mL, which is 1/10 of a Liter, we also only need 1/10 the amount of NaCl, or 292.2 / 10, which is 29.22g. So, measure out 29.22g NaCl, and dissolve completly in a volume less than 100mL, say 80mL, then bring the final volume up to 100mL. You now have 100mL of a 5M 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.
calculate the osmolarity of a solution ; 1m of sucrose at 25 degree centigra 2m kcl at 25degree centigre 154mM Nacl at 25 degree
The formula weight is 121.5 --> this is equivalent to 1M with 121.5g tris in 1L dH20. For a 5M stock, use 5x as much tris in the same 1L dh20.607.5 g tris into 800ml dH2O - stirring - then pH to 7.5 with 6M HCl and QS to your final volume of 1L
2g
1.3g
0,4 mol NaCl is 23,376 g.2,85 mol NaCl is 166,554 g to 1L.140 mL solution NaCl 2,85 M contain 0,4 mol NaCl.
Gram percent is the number of grams of a solute per 100 grams of a solution. For example, if a solution of NaCl and water was said to have a 0.02g% of NaCl, this would mean that for 100g of saline solution, 0.02 of those grams are salt. Since 1L of water weighs 1kg (at normal conditions), there would be .2g of NaCl in 1L of a 0.02g% NaCl solution.
The Molecular Weight of NaCl = 58.5 So to make 1L of 4M NaCl solution you need 4*58.5=234g of NaCl So to make 100mL of the above solution you need 23.4 grams of NaCl
V = 141.37 m3
calculate final molarity of the solution if 11ml of 5m solution is made up to 20ml
26328