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
xx mmol x 1 mole/1000 mmol x 58.5 g/mole
Write an equation representing the ppm concentration: ppm = mass solute (mg) ÷ volume solution (L) Extract the data from the question: mass solute (NaCl) = 0.0045 g. ... Convert the mass in grams to a mass in milligrams: mass NaCl = 0.0045 g = 0.0045 g × 1000 mg/g = 4.5 mg.
It means you dilute your sample in a volume that is as great as the one you current sample has. Ex: you dilute 50 ml NaCl-solution in 50 ml MQ-water. The result of this is that the concentration will always be halved, seeing as the volume increases twofold.
No. Usually, the second compound is negitive.
Abstract Envelopes of a marine isolate, c-A1, and of a terrestrial isolate, 121, were compared for their susceptibility to disintegration in distilled water after exposure to 0.05 m MgCl2 and to 0.1 and 1.0 m NaCl. After exposure to MgCl2 alone, both types of envelopes remained intact in distilled water. Envelopes of marine isolate c-A1, but not of the terrestrial isolate, fragmented in distilled water after exposure to 1.0 m NaCl. Partial reaggregation of the c-A1 envelope fragments occurred on addition of MgCl2. In cation-exchange experiments, bound Mg++ in the envelopes of both organisms was displaced by Na+. The envelopes of c-A1 were found to contain lipopolysaccharide, muramic acid, and a variety of phospholipids, of which the major component was phosphatidylethanolamine, accompanied by lesser amounts of phosphatidic acid, diphosphatidylglycerol, and phosphatidylserine. Analyses of envelope acid hydrolysates revealed a similar amino acid distribution in the marine and terrestrial isolates, but envelopes of c-A1 had less than half the total amino acid content of envelopes of 121 per envelope dry weight. Possible relationships between cations and biochemical components of the envelopes are considered in terms of differences in behavior of the two organisms in low ionic environments.
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).
To prepare a 6M NaCl solution, you would need to dissolve 58.44 grams of NaCl in water to make a 1 liter solution. To prepare a different volume, you would adjust the amount of NaCl accordingly using the molarity equation: Molarity = moles of solute / volume of solution in liters.
The answer is 0,9 g pure, dried 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.
The concentration of NaCl in a solution that contains 0.9 NaCl is 0.9 grams of NaCl per 100 grams of solution.
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.
To completely neutralize 100ml of 1M H2SO4, you would need an equal number of moles of NaCl. H2SO4 is a diprotic acid, so it will require 2 moles of NaCl to neutralize 1 mole of H2SO4. Therefore, you would need 2 moles of NaCl for every mole of H2SO4. With a 1M solution of H2SO4 in 100ml, you have 0.1 moles of H2SO4. Therefore, you would need 0.2 moles of NaCl. The molar mass of NaCl is approximately 58.44g/mol, so you would need approximately 11.7 grams of NaCl to completely neutralize the 1M H2SO4 solution.
To prepare a Normal Saline Solution in 100ml of distilled water, you would add 0.9g of sodium chloride (NaCl) to the water. This concentration of 0.9% NaCl is considered isotonic and mimics the salt concentration of our body fluids. Stir the solution until the salt is completely dissolved before use.
To calculate the weight of NaCl in grams, you need to know the concentration of NaCl in the solution or material. Without this information, it's not possible to determine the weight of NaCl in grams.
NaCl doesn't neutralize sulfuric acid.
To prepare 100 ml of 1000 ppm Na solution from NaCl, first calculate the amount of NaCl needed based on its molecular weight. Then dissolve this amount in water to make a 100 ml solution. Since 1000 ppm means there are 1000 mg of Na per liter of solution, you would need to add 10 mg of NaCl to achieve this concentration.
The answer is 8 g NaCl.