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
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.
xx mmol x 1 mole/1000 mmol x 58.5 g/mole
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.
58 grams of NaCl in cylinder measure water to 100ml
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.
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.
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.
NaCl doesn't neutralize sulfuric acid.
The answer is 8 g NaCl.
That refers to a mixture consisting of 2/100 of sodium chloride (salt) and 98/100 of something else (usually water).
Put 0.9g of Nacl in a beaker and add distilled water to make up to 100ml. that is 0.9% Nacl (Normal saline Solution) From Tade Olubunmi Ademola
You need 841,536 g NaCl.
117 grams of NaCl
If your solution is a total of 414g and 3.06% of it needs to be NaCl, then you just take 414 x .0306 = grams of NaCl. The rest of the grams will be from other species in the solution.