Search results
Results from the WOW.Com Content Network
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 ...
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:
The solution has 1 mole or 1 equiv Na +, 1 mole or 2 equiv Ca 2+, and 3 mole or 3 equiv Cl −. An earlier definition, used especially for chemical elements , holds that an equivalent is the amount of a substance that will react with 1 g (0.035 oz) of hydrogen , 8 g (0.28 oz) of oxygen , or 35.5 g (1.25 oz) of chlorine —or that will displace ...
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.
The molar mass of atoms of an element is given by the relative atomic mass of the element multiplied by the molar mass constant, M u ≈ 1.000 000 × 10 −3 kg/mol ≈ 1 g/mol. For normal samples from Earth with typical isotope composition, the atomic weight can be approximated by the standard atomic weight [ 2 ] or the conventional atomic weight.
The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...
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.
For example, the conversion factor between a mass fraction of 1 ppb and a mole fraction of 1 ppb is about 4.7 for the greenhouse gas CFC-11 in air (Molar mass of CFC-11 / Mean molar mass of air = 137.368 / 28.97 = 4.74). For volume fraction, the suffix "V" or "v" is sometimes appended to the parts-per notation (e.g. ppmV, ppbv, pptv).