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The following formulas can be used to calculate the volumes of solute (V solute) and solvent (V solvent) to be used: [1] = = where V total is the desired total volume, and F is the desired dilution factor number (the number in the position of F if expressed as "1/F dilution factor" or "xF dilution"). However, some solutions and mixtures take up ...
For example, if there are 10 grams of salt (the solute) dissolved in 1 litre of water (the solvent), this solution has a certain salt concentration . If one adds 1 litre of water to this solution, the salt concentration is reduced. The diluted solution still contains 10 grams of salt (0.171 moles of NaCl).
m(NaCl) = 2 mol/L × 0.1 L × 58 g/mol = 11.6 g. To create the solution, 11.6 g NaCl is placed in a volumetric flask, dissolved in some water, then followed by the addition of more water until the total volume reaches 100 mL. The density of water is approximately 1000 g/L and its molar mass is 18.02 g/mol (or 1/18.02 = 0.055 mol/g). Therefore ...
The survival ratio is commonly expressed as an inactivation ratio (in %) or as the number of reductions in the order of magnitude of the microorganism concentration. For example, a situation where N 0 =10 7 CFU /L and N=10 4 CFU/L would be reported as a 99.9% inactivation or "3-log 10 " removal.
If one begins with a solution of 1 mol/L of a substance, the dilution required to reduce the number of molecules to less than one per litre is 1 part in 1×10 24 (24X or 12C) since: 6.02×10 23 /1×10 24 = 0.6 molecules per litre
Osmotic concentration, formerly known as osmolarity, [1] is the measure of solute concentration, defined as the number of osmoles (Osm) of solute per litre (L) of solution (osmol/L or Osm/L). The osmolarity of a solution is usually expressed as Osm/L (pronounced "osmolar"), in the same way that the molarity of a solution is expressed as "M ...
Since only 0.5 mol of H 2 SO 4 are needed to neutralize 1 mol of OH −, the equivalence factor is: f eq (H 2 SO 4) = 0.5. If the concentration of a sulfuric acid solution is c(H 2 SO 4) = 1 mol/L, then its normality is 2 N. It can also be called a "2 normal" solution.
One can calculate the amount of AgCl that will dissolve in 1 liter of pure water as follows: K sp = [Ag +] × [Cl −] / M 2 (definition of solubility product; M = mol/L) K sp = 1.8 × 10 −10 (from a table of solubility products) [Ag +] = [Cl −], in the absence of other silver or chloride salts, so [Ag +] 2 = 1.8 × 10 −10 M 2 [Ag +] = 1. ...