Question

1.22g of benzoic acid is dissolved in 100g of acetone (K on a base 2.6K Kg mol inverse) and 100g of benzene (K on a base=2. 6 k Kg mol inverse). If the elevation in boiling point ∆T base is 0.26°C and 0.13°C respectively

Answer 1

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Answer 2

Although your question is incomplete, you may be referring to the full question:

1.22g of benzoic acid is dissolved in 100g of acetone (Kb= 2.6K Kg mol inverse) and 100g of benzene (Kb=2. 6 k Kg mol inverse). If the elevation in boiling point ∆T base is 0.26°C and 0.13°C respectively, what is the molar mass of benzoic acid in the two solvents?

Given: 1.22g of benzoic acid is dissolved in 100g of acetone (Kb= 2.6K Kg mol inverse) and 100g of benzene (Kb =2. 6 k Kg mol inverse). Elevation in boiling points is 0.26°C and 0.13°C respectively.

To find: Molar mass of benzoic acid in the two solvents

Solution: Let us divide the question in two cases:

Case-I: When benzoic acid is dissolved in acetone

Here, benzoic acid is the solute and acetone is the solvent.

Mass of acetone= 100 g = 0.1 kg

Mass of benzoic acid(m)= 1.22 g

Molar mass of benzoic acid = 122 g

Therefore, moles of benzoic acid

= Mass / molar mass

= 1.22/122

= 0.01

Molality of solution(m)

= Moles of solute/ Mass of solvent

= 0.01/0.1

= 0.1

Rise in temperature in acetone= 0.13°C

Kb = 2.6

Now, rise in temperature (∆T) is given by:

∆T = Kb × m × i where i is Vant's Hoff value

Using the values:

0.26 = 2.6 × 0.1 × i

=> 0.26 = 0.26 × i

=> i = 1

i = Original Molar mass of benzoic acid/ Molar mass of benzoic acid in solution(acetone)

=> 1 = 122 / Molar mass in acetone

=> Molar mass in acetone = 122 g/mol

Case-II: When benzoic acid is dissolved in benzene

Here, benzoic acid is the solute and benzene is the solvent.

Mass of benzene= 100 g = 0.1 kg

Mass of benzoic acid(m)= 1.22 g

Molar mass of benzoic acid = 122 g

Therefore, moles of benzoic acid

= Mass / molar mass

= 1.22/122

= 0.01

Molality of solution(m)

= Moles of solute/ Mass of solvent

= 0.01/0.1

= 0.1

Rise in temperature in acetone= 0.26°C

Kb = 2.6

Now, rise in temperature (∆T) is given by:

∆T = Kb × m × i where i is Vant's Hoff value

Using the values:

0.13 = 2.6 × 0.1 × i

=> 0.13 = 0.26 × i

=> i = 0.13/0.26

=> i = 0.5

i = Original Molar mass of benzoic acid / Molar mass of benzoic acid in solution(benzene)

=> 0.5 = 122 / Molar mass in benzene

=> Molar mass in benzene = 122/ 0.5

=>Molar mass in benzene = 244 g/mol

Therefore, the molar mass of benzoic acid in acetone and benzene is 122 g/mol and 244g/mol respectively.