Dissolve 100 mg NaCl in 1.0 Litre water.
It depends on the final solution Volume you want to prepare. For 100ml of a 6M NaCL solution, you add 35.1g of NaCl to water until you reach 100ml. Dissolve and autoclave for 15 mins.
NaCl is the formula for Sodium Chloride, from that you can work out the gram formula mass. RAM Na = 23 RAM Cl = 35.5 gfm (NaCl) = 58.5g mol-1 assuming that you need 0.1l of NaCl m = 58.5*0.1 = 5.85g So dissolve 5.85g in water for a 0.2 mole solution
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.
There would be 0.1 moles of NaCl present in 1 liter of a 0.1M solution of sodium chloride. This is based on the definition of molarity which is moles of solute per liter of solution.
Oh honey, it's not rocket science. Just measure out 30 grams of table salt and dissolve it in enough water to make 100 grams of solution. Voila, you've got yourself a 30% NaCl solution. Just don't go drinking it thinking it's a margarita, okay?
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
This is a solution containing 100 mg NaCl/L.
It depends on the final solution Volume you want to prepare. For 100ml of a 6M NaCL solution, you add 35.1g of NaCl to water until you reach 100ml. Dissolve and autoclave for 15 mins.
NaCl is the formula for Sodium Chloride, from that you can work out the gram formula mass. RAM Na = 23 RAM Cl = 35.5 gfm (NaCl) = 58.5g mol-1 assuming that you need 0.1l of NaCl m = 58.5*0.1 = 5.85g So dissolve 5.85g in water for a 0.2 mole solution
What volume of this solution do you desire? Let's say you want to make 1 liter of such a solution. You would weigh out 1 gram (1000 mg) of NaCl and dissolve it in enough water to make a final volume of 1 liter (1000 ml). Since 1000 ppm means 1000 mg/liter, this is how you make 1 liter of that solution. For larger or smaller volumes, adjust appropriately.
Molar mass of NaCl =~58.4 g/mole0.1 N NaCl = 0.1 moles/liter To make 1 liter of 0.1N NaCl thus requires 0.1 moles/liter = 0.1 moles x 58.4 g/mole = 5.84 moles Dissolve 5.84 g (6 g using 1 sig. fig.) in a final volume of 1 liter to make 0.1N NaCl
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.
To prepare a 100 mM NaCl solution, you would need to calculate the molecular weight of NaCl, which is approximately 58.44 g/mol (sodium's atomic weight is 22.99 g/mol and chlorine's is 35.45 g/mol). To make a 100 mM solution, you would need 0.1 moles of NaCl per liter of solution. This would be equivalent to 5.844 grams of NaCl per liter of solution.
0.5 M means 0.5 moles per liter. so it depends on how many liters of solution that you need.Suppose you want to make 1 liter of solution, then you need 0.5 moles of NaClFrom the webelements.com Periodic Table:Atomic mass of Sodium (Na) = 22.990 & Atomic Mass of Chlorine (Cl) = 35.453So 1 mole of NaCl = (22.990 + 35.453) grams = 58.443 gramsBased on the 1 liter, we want to add 0.5 moles: (58.443 grams/mole)*(0.5 mole) = 29.2215 g (to make 1 liter of 0.5 M solution)
To make 1 liter of 0.1 M NaCl solution, you will need 25 ml of the 4 M NaCl stock solution and 975 ml of water. This will give you the desired concentration of 0.1 M NaCl in a total volume of 1 liter.
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.
There would be 0.1 moles of NaCl present in 1 liter of a 0.1M solution of sodium chloride. This is based on the definition of molarity which is moles of solute per liter of solution.