Question

47.) 6.055 x 10-2 kg of washing soda (Na2co3.10h2o) is dissolved in water to obtain 1 L of a solution of density
1077 .2 kg/m?. Calculate the molarity, molality and mole fraction of Na2CO3 in the solution.

Answer 1

Answer:

Refer to the attachment for the answer.

Answer 2

Concept:

Molarity is one of the most often used concentration units and is represented by the letter M. Number of moles of solute present in 1 liter of solution is how it is defined.

Molality is described as the number of moles of solute in one kilogram of solvent. m designates it.

The quantity of a constituent (represented in moles), ni, divided by the sum of all the constituent amounts in a mixture is known as the mole fraction.

Given:

The weight of washing soda = 6.055 x 10⁻² Kg = 6.055 x 10⁻² x 10³ g = 60.55 g

The volume of water = 1 L = 10³ cm³

The density of the solution = 1077.2 kg/m³ = 1077.2 x 10³/10⁶ g/cm³ = 1077.2 x 10⁻³ g/cm³

Find:

Calculate the molarity, molality, and mole fraction of Na₂CO₃ in the solution.

Solution:

The molecular weight of Na₂CO₃.10H₂O = 286 g/mol

The molecular weight of Na₂CO₃ = 106 g/mol

The density of solution = Mass of solution / Volume of solution

Mass of solution = Density x Volume

Mass of solution = 1077.2 x 10⁻³ g/cm³ x 10³ cm³

Mass of solution = 1077.2 g

Thus, Mass of solvent = Mass of solution - Mass of solute

= 1077.2 g - 60.55 g

= 1016.65 g

Moles of Na₂CO₃.10H₂O = Mass/Molar mass = 60.55 g / 286 g/mol = 0.21 mol

Moles of Na₂CO₃ = 60.55 g / 106 g/mol = 0.57 mol

Moles of H₂O = 1016.65 g / 18 g/mol = 56.48 mol

Molality = Moles of solute / Volume of solvent (in Kg) = 0.21 mol / 1.01665 Kg = 0.21 mol/Kg

Molarity = Moles of solute / Volume of solution (in liters) = 0.21 mol / 1 L = 0.21 mol/L = 0.21 M

Mole fraction of Na₂CO₃ = Number of moles of Na₂CO₃ / Number of moles of(Na₂CO₃ + H₂O)

= 0.57 / (0.57+56.4)

= 0.57 / 56.97

= 0.01

Hence, the molarity, molality, and mole fraction of Na₂CO₃ in the solution are 0.21 M, 0.21 mol/Kg, and 0.01 respectively.

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