Question

A solution containing 0.5126 g of naphthalene (molar mass= 128.17 g mol⁻¹) in 50.00 g of CCl₄ gives a boiling point elevation of 0.402 K. While a solution of 0.6216 g of unknown solute in the same mass of the solvent gives a boiling point elevation of 0.647 K. Find the molar mass of the unknown solute. (Kb for CCl₄ = 5.03 K kg mol⁻¹ of solvent) (96.7 g mol⁻¹)

Answer 1

Answer:

The molar mass of the unknown solute, M2’ is 96.7 g/mol.

Explanation:

Given data:

Weight of CCl4, W1 = 50.00 g

Weight of naphthalene, W2 = 0.5126 g

Molar mass of naphthalene, M2 = 128.17 g mol⁻¹

Kb for CCl4  = 5.03 K kg mol⁻¹ of solvent

Boiling point elevation, ∆Tb = 0.402 K

Weight of unknown solute, W2’ = 0.6216 g

Weight of unknown solvent, W1’ = 50 g

Boiling of elevation in case of the unknown solution, ∆Tb’ = 0.647 K

To find: the molar mass of the unknown solute, M2’

The general equation of ∆Tb is given by  

∆Tb = Kb * m  

Where, Kb = ebullioscopic constant and m = molality of the solute

Therefore, for the unknown solute, we can write the equation as  

∆Tb’ = Kb * m’ = Kb * [$\frac{W2'}{W1 * M2'}$]

Or,  0.647 = 5.03 * 1000 *  [(0.6216 / 50) * (1/M2’)]

Or, M2’ = $\frac{5.03 * 12.432}{0.647}$

Or, M2’ = $\frac{62.532}{0.647}$ = 96.65 g/mol ≈ 96.7 g/mol.

Answer 2

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