Mole Fraction Formula
Mole Fraction Formula
Mole Fraction Formula is a unit of concentration in chemistry. One Mole Fraction Formula is actually a variety of molecules, approximately 6 × 1023. A Mole Fraction Formula is the ratio of molecules of 1 component in a mixture. If the Mole Fraction Formula of the methane gas is 0.90, then this means that 90% of the molecules in the mixture are methane. Since the quantity fractions are very similar to the mole fractions, the mixture is additionally 90%, methane and the Mole Fraction Formula helps calculate these numbers. This article discusses the Mole Fraction Formula, all its properties, and solves examples.
What is Mole fraction?
The mole fraction of all solutes is defined as the number of moles of that solute divided by the total number of moles of all solutes and the solvent.
Mole Fraction Formula is just like the amount fraction. Instead of a Roman X, the lowercase Greek letter (chi) is used to represent a molal fraction. For mixtures of gases, IUPAC recommends the letter “y” to be confused because it helps with better comprehension. The Mole Fraction Formula is a method of expressing the composition of a mixture, which is essentially a dimensionless quantity, as the mass fraction (percentage by weight, wt%) and volume fraction (percentage by volume, vol%).
The advantage of using the dimensionless mole fraction scale in some problems is that when there is a dearth of reactions, the mole fraction of chemical substance k is independent of both temperature and pressure, meaning that these factors do not influence the reaction or the number at all. The utilisation of the mole fraction allows the calculation to be done for various mixtures of gases, which in a traditional sense would have been difficult.
Properties of Mole Fraction
Mole Fraction Formula is put to use very frequently within the construction of phase diagrams. The Mole Fraction Formula is not at all dependent on the temperature of the substance (such as molar concentration), and it absolutely doesn’t require knowledge of the densities of the phases involved in the entire process. A mix of known mole fractions is often prepared by weighing off the acceptable masses of the constituents to make the process complete with great ease.
The measure of the Mole Fraction Formula is symmetric:
Within the mole fractions, x = 0.1 and x = 0.9, the roles of ‘solvent’ and ‘solute’ are reversed.
In the case of an ideal gas mixture, the mole fraction is frequently expressed in terms of the ratio of the partial pressure to the total pressure of the mixture. When it is a ternary mixture, one can express mole fractions of a component as functions of other components’ mole fractions and binary mole ratios.
Mole fraction is very useful when two reactive components are mixed together because the ratio of the 2 components can be understood if the mole fraction of each is known. Multiplying the mole fraction by 100 yields a mole percent, which is an important quantity to consider in the case of the Mole Fraction Formula, which describes the same thing as the mole fraction but in a different form.
Solved Examples for Mole Fraction Formula
A Solution contains 30% by mass in carbon tet. Calculate the mole fraction of benzene using the Mole Fraction Formula.
The total mass of the substance concerned = 100 g
The mass of benzene = 30 g
Therefore, the mass of carbon tet = (100 – 30)g
Benzene’s molar mass (C6H6) = (6 × 12 + 6 × 1) g mol – 1 = 78 g mol – 1
Hence, C6H6’s number of moles =30/78 mol = 0.3846 mol
Molar mass of carbon tet (CCl4) = 1 × 12 + 4 × 355 = 154 g mol – 1
∴ CCl4’s number of moles = 70/154 mol = 0.4545 mol
Mole fraction of C6H6 = [number of moles of C6H6 / number of moles of C6H6 + number of moles of (CCl4) ] X 100%
= 0.3846/ 0.3846+ 0.4545 = 0.458
Students must take care of this subject with great precision, primarily because the whole chapter concerns mathematical problems, but the concepts are crucial to be understood if students desire a smooth transition between the chapters. If students are not careful while they are learning about these ideas, then they will make very bad errors in the exam, losing a lot of marks.
