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Normality is defined as the number of gram or mole equivalents of solute present in one liter of solution.The SI unit of normality is equivalents per liter (Eq/L). = where N is normality, m sol is the mass of solute in grams, EW sol is the equivalent weight of solute, and V soln is the volume of the entire solution in liters.
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 ...
This page lists examples of the orders of magnitude of molar concentration. Source values are parenthesized where unit conversions were performed. M denotes the non-SI unit molar: 1 M = 1 mol/L = 10 −3 mol/m 3.
Sometimes specific volume is expressed in terms of the number of cubic centimeters occupied by one gram of a substance. In this case, the unit is the centimeter cubed per gram (cm 3 /g or cm 3 ·g −1). To convert m 3 /kg to cm 3 /g, multiply by 1000; conversely, multiply by 0.001. Specific volume is inversely proportional to density.
With this conversion from SCCM to kg/s, one can then use available unit calculators to convert kg/s to other units, [5] such as g/s of the CGS system, or slug/s. Based on the above formulas, the relationship between SCCM and molar flow rate in kmol/s is given by
A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). This is incorrect because the unit "%" can only be used for dimensionless quantities. Instead, the concentration should simply be given in units of g/mL.
0.0248 * (1000 g/L * (mol/18.0153g)) is the molar fraction of substance in saturated solution with a unit of mol/L; 0.0248 * (1000 g/L * (mol/18.0153g)) * 151.17g/mol is the solute's molar fraction equivalent mass conversion; 1-0.0248 will be the fraction of the solution that is solvent. which is a deviation from the real solubility (240 g/L ...
V eq is the volume of titrant (ml) consumed by the crude oil sample and 1 ml of spiking solution at the equivalent point, b eq is the volume of titrant (ml) consumed by 1 ml of spiking solution at the equivalent point, 56.1 g/mol is the molecular weight of KOH, W oil is the mass of the sample in grams. The normality (N) of titrant is calculated as: