Calculate the boiling point of solution prepared by adding 2.93 g of solute to 0.059 kg of solvent, if the boiling point of pure solvent = 61. 2 ∘ C , K b = 3.8 5 ∘ C Kg mol − 1 and molar mass of so

Question

Calculate the boiling point of solution prepared by adding 2.93 g of solute to 0.059 kg of solvent, if the boiling point of pure solvent = 61. 2 ∘ C , K b ​ = 3.8 5 ∘ C Kg mol − 1 and molar mass of solute = 32 g/mol)

TestHub · Accepted Answer

The elevation in boiling point is given by the formula:

$Δ T_{b} = K_{b} \times m$

where $m$ is the molality of the solution.

Molality $m = \frac{w _{2}}{M _{2} \times w _{1}}$

Given:
$w_{2} = 2.93$ g
$M_{2} = 32$ g/mol
$w_{1} = 0.059$ kg
$K_{b} = 3.8 5^{\circ}$ C kg mol $^{- 1}$

Substituting the values:

$m = \frac{2.93}{32 \times 0.059} = \frac{2.93}{1.888} = 1.552$ mol kg $^{- 1}$

$Δ T_{b} = 3.85 \times 1.552 = 5.9 7^{\circ}$ C

The boiling point of the solution $T_{b}$ is:

$T_{b} = T_{b}^{\circ} + Δ T_{b}$

$T_{b} = 61. 2^{\circ}$ C $+ 5.9 7^{\circ}$ C $= 67.1 7^{\circ}$ C

Answer: $67.17^{0} C$

Chemistry - Liquid Solution Question with Solution | TestHub

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