 What is the relation between mole fraction and molality?

The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all the components of the solution.

$$Mole fraction =\frac{Number of moles of solute}{Number of moles of solvent}$$ $$mole fraction = \frac{n_{B}}{n_{A}+n_{B}}$$

nB= number of moles of solute

nA = Number of moles of solvent

Molality is defined as the total moles of a solute contained in a kilogram of a solvent.

$$Molality = \frac{Moles of solute}{Moles of solven}$$ $$Molality = \frac{n_{B}}{W_{A}(in kg)}$$

Solution

The relation between the mole fraction and molality is given as:

Consider a binary solution component with A as solvent and B as solute.

Let,

Mole fraction of solvent = xA

Mole fraction of solute = xB

Number of moles of solvent = nA

Number of moles of solute = nB

Mass of solvent = WA

Mass of solute = WB

Molar mass of solvent = MA

The molar mass of solute = MB

Mole fraction of solute = xB = $$\frac{n_{B}}{n_{B}+n_{A}}$$ (eq.1)

Mole fraction of solvent = xA = $$\frac{n_{A}}{n_{B}+n_{A}}$$ (eq.2)

Dividing eq 1 and 2,

$$\frac{X_{B}}{X_{A}} = \frac{\frac{n_{B}}{n_{B}+n_{A}}}{\frac{n_{A}}{n_{B}+n_{A}}}$$

Therefore,

$$frac{x_{B}}{x_{A}}=frac{frac{W_{B}}{M_{B}}}{frac{W_{A}}{M_{A}}}=frac{W_{B}}{W_{A}}times frac{M_{A}}{M_{B}}$$

Therefore,

$$frac{x_{B}times 1000}{x_{A}times M_{A}}=frac{W_B}{M_B}times frac{1000}{W_A}=frac{n_Btimes 1000}{W_A}$$ = Molality = m

Therefore,

Molality = m = $$frac{x_{B}times 1000}{x_{A}times M_{A}}=frac{x_Btimes 1000}{(1-x_B)times M_A}$$

Solved Examples

1. Calculate the mole fraction of ammonia in a 3.00 m solution of ammonia in water.

Solution:

Let us assume 1 kg of solvent that is 1000 g of water

The molecular mass of water = 18.015 g mol1

Moles of water = 1000/18.015

Moles of water = 55.402

For 1 kg of H2O, there are 3 moles of NH3

$$Mole fraction = frac{number of moles of solute}{Number of moles of solvent}$$ $$Mole fraction = frac{3}{3 + 55.402}$$

Mole freaction = 0.0514

2.  Calculate the molality of the solution when 70.128 grams of NaCl salt is dissolved in 2 grams of water.

Solution:

Number of moles of NaCl = 70.128 g/ 58.44

Molar mass of NaCl = 58.44 g/mol

Number of moles of NaCl = 1.2 moles

Molality = 1.2/2

Molality = 0.6 mol/kg

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