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In chemistry, the molar mass (M) (sometimes called molecular weight or formula weight, but see related quantities for usage) of a chemical compound is defined as the ratio between the mass and the amount of substance (measured in moles) of any sample of the compound. [1] The molar mass is a bulk, not molecular, property of a substance.
M is the molar mass of the solvent. T b is boiling point of the pure solvent in kelvin. ΔH vap is the molar enthalpy of vaporization of the solvent. Through the procedure called ebullioscopy, a known constant can be used to calculate an unknown molar mass. The term ebullioscopy means "boiling measurement" in Latin.
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 ...
This template calculates the molecular mass (or molar mass) of a chemical compound. It is designed to be embedded in infoboxes {{ Infobox drug }} and {{ Chembox }} , but it can be used in-line just as well.
The number average molar mass is a way of determining the molecular mass of a polymer.Polymer molecules, even ones of the same type, come in different sizes (chain lengths, for linear polymers), so the average molecular mass will depend on the method of averaging.
The molar mass constant, usually denoted by M u, is a physical constant defined as one twelfth of the molar mass of carbon-12: M u = M(12 C)/12. [1] The molar mass of an element or compound is its relative atomic mass (atomic weight) or relative molecular mass (molecular weight or formula weight) multiplied by the molar mass constant.
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M 2 is the molar mass of gas 2. Graham's law states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. Thus, if the molecular weight of one gas is four times that of another, it would diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate ...