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#### Define Mole Fraction And Mole Percent: Explain How Of Its Can Be Calculated

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### Define mole fraction and mole percent.

In order to understand the concept of mole fraction and mole percent, the concept of mole is needed.

Mole:The mole of a substance is defined as the mass of substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of 12 C.That number of particles is Avogadro's Number, which is roughly 6.02x1023. A mole of carbon atoms is 6.02x1023 carbon atoms.

Mole fraction:Mole fraction is defined as the ratio of moles of a particular component to the total number of moles of a solution.Mole fraction is the ratio of moles of a particular component to the total number of moles of a solution.

Mole fractions are dimensionless, and the sum of all mole fractions in a given mixture is always equal to 1.

nt=na+nb+nc+… , where

nt = total number of moles

na = moles of component a

nb = moles of component b

If a substance ‘a’ dissolves in substance 'b' and their number of moles are na and nbrespectively; then the mole fractions of a and b are given as:

i) Mole fraction of a (χa)

=(no. of moles of A )÷(No. of moles of solution)

=na÷ ( na + nb)

ii) Mole fraction of b (χb)

=(No. of moles of B )÷(no. of moles of A + no. of moles of B)

=nb ÷ (na + nb)

### Properties of the Mole Fraction

The mole fraction is used very frequently in the construction of phase diagrams. It has a number of advantages:

i) It is not temperature dependent, as opposed to molar concentration, and does not require knowledge of the densities of the phase(s) involved.

ii) A mixture of known mole fractions can be prepared by weighing the appropriate masses of the constituents.

iii) The measure is symmetric; in the mole fractions x=0.1 and x=0.9, the roles of ‘solvent’ and ‘solute’ are reversible.

iv) In a mixture of ideal gases, the mole fraction can be expressed as the ratio of partial pressure to total pressure of the mixture.

Example 1:A solution is prepared by mixing 25.0 g of water, H2O, and 25.0 g of ethanol, C2H5OH. Determine the mole fractions of each substance.

Solution:

i) Determine the moles of each substances:

H2O ⇒ 25.0 g / 18.0 g/mol = 1.34 mol

C2H5OH ⇒ 25.0 g / 46.07 g/mol = 0.543 mol

ii) Determine the mole fraction:

H2O ⇒ 1.34 mol/ (1.34 mol + 0.543 mol) = 0.71

C2H5OH ⇒0.543 mol / (1.34 mol + 0.543 mol) = 0.29

Mole Percent:Multiplying the mole fraction by 100 gives the mole percentage, also referred as amount/amount percent (abbreviated as n/n%). For general chemistry, all the mole percents of a mixture add up to 100 mole percent. In order to convert again mole percent back to mole fraction, the mole percent should be divided by 100.

mole % a = χa × 100 %

The sum of the mole percentages for each component in a solution is equal to 100 %.

For a solution containing two components, solute and solvent,

mole % solute + mole % solvent = 100 %

### Explain how both of these can be calculated, using a suitable example.

Example 1:

What are the mole fraction and mole percent of sodium chloride and the mole fraction and mole percent of water in an aqueous solution containing 25.0 g water and 5.0 g sodium chloride?

Solution:

i) Identify the components making up the solution.

NaCl is the solute and H2O is the solvent.

ii) Calculate the moles of each component. iii) Calculate the mole fraction of each component.

nt = n1 + n2 = ( 1.38 + 0.086      ) mol = 1.46 mol However, it could have also be written as χ1=1–χ2=1–0.058=0.942

iv)        Calculate the mole percent of each component.

Mole % 2 = χ2 × 100% = 0.058 × 100% = 5.8 mole %

Mole % 1 = χ1 × 100% = 0.942 × 100% = 94.2 mole %

[Note: it could have also be written as mole % 1 = 100 – mole % 2 = (100 – 5.8) % = 94.2 mole %]

In short the summary of the solution is given as: Example 2 :

Find the mole fraction of Methanol ( CH3OH ) and water in a solution prepared by dissolving 4.5 g of alcohol in 40 g of H2O. Also determine the mole percent of Methanol ( CH3OH ). Molar Mass of H2O is 18 gm / mole and Molar mass of CH3OH is 32 gm / mole.

Solution :

Moles of CH3OH = 4.5 / 32 = 0.14 mole

Moles of H2O = 40 / 18 = 2.2 moles

Therefore, according to the equation

Mole fraction of CH3OH = 0.14 / 2.2 + 0.14

= 0.061

Mole fraction of H2O. = 2.2 / 2.2 + 0.14

= 0.940

The Mole percent of H2O. = 0.940 × 100 %

= 94 mole %

Example 3 :

Calculate the mole fraction of HCl and H2O in a solution of hydrochloric acid in water, containing 30% HCl by weight. Also determine the mole percent.

Solution :

The solution contains 30 grams of Hydrochloric acid and 70 grams of water.

Also,

Molar mass of HCl is 36.5 grams / mole.

Molar mass of water is 18 grams / mole.

Moles of HCl = 30 / 36.5 = 0.821 moles of HCl

Moles of water = 70 / 18 = 3.88 moles of H2O.

Mole fraction of HCl = 0.821 / ( 0.821 + 3.88 ) = 0.174

Mole fraction of H2O = 3.88 / ( 0.821 + 3.88 ) = 0.825

Mole percent of HCl = 0.174 × 100 = 17.4 mole %

Mole percent of H2O = 0.825 × 100 = 82.5 mole %

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