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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 ...
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 molar concentration of water is
The mole is widely used in chemistry as a convenient way to express amounts of reactants and amounts of products of chemical reactions. For example, the chemical equation 2 H 2 + O 2 → 2 H 2 O can be interpreted to mean that for each 2 mol molecular hydrogen (H 2) and 1 mol molecular oxygen (O 2) that react, 2 mol of water (H 2 O) form.
Thus 100 mL of water is equal to approximately 100 g. Therefore, a solution with 1 g of solute dissolved in final volume of 100 mL aqueous solution may also be considered 1% m/m (1 g solute in 99 g water). This approximation breaks down as the solute concentration is increased (for example, in water–NaCl mixtures). High solute concentrations ...
When one mole of water is added to a large volume of water at 25 °C, the volume increases by 18 cm 3. The molar volume of pure water would thus be reported as 18 cm 3 mol −1. However, addition of one mole of water to a large volume of pure ethanol results in an increase in volume of only 14 cm 3. The reason that the increase is different is ...
The volume of such a mixture is slightly less than the sum of the volumes of the components. Thus, by the above definition, the term "40% alcohol by volume" refers to a mixture of 40 volume units of ethanol with enough water to make a final volume of 100 units, rather than a mixture of 40 units of ethanol with 60 units of water.
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). Mathematically this relationship can be shown by equation: = where c 1 = initial concentration or molarity; V 1 = initial volume
Historically, the mole was defined as the amount of substance in 12 grams of the carbon-12 isotope.As a consequence, the mass of one mole of a chemical compound, in grams, is numerically equal (for all practical purposes) to the mass of one molecule or formula unit of the compound, in daltons, and the molar mass of an isotope in grams per mole is approximately equal to the mass number ...