If the boiling point elevation constant and the boiling point of benzene are 2-16 K kg mol-1 and 81-0- C respectively- calculate the boiling point of a solution formed when 5 g of C14 H12 is dissolved in 15 g of benzene-

Δ T_b = k_b × m molality = number of moles of solute/weight of solvent× 1000 Molar mass of C_14H_12 = 180 g/mol molality = 5/180/15× 1000 = 1.85 m Δ T_b = k_b ...

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Byju's Answer

Standard XIII

Chemistry

Elevation in Boiling Point

If the boilin...

Question

If the boiling point elevation constant and the boiling point of benzene are $2.16 {K kg mol}^{- 1}$ and ${81.0}^{\circ} C$ respectively, calculate the boiling point of a solution formed when $5 g$ of $C_{14} H_{12}$ is dissolved in $15 g$ of benzene.

90^{\circ} C

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75^{\circ} C

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81^{\circ} C

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85^{\circ} C

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Solution

The correct option is D $85^{\circ} C$
$Δ T_{b} = k_{b} \times m$
$m o l a l i t y = \frac{number of moles of solute}{weight of solvent} \times 1000$
Molar mass of $C_{14} H_{12}$ = 180 g/mol
$m o l a l i t y = \frac{5 / 180}{15} \times 1000 = 1.85 m$
$Δ T_{b} = k_{b} \times m$
$= 2.16 \times 1.85 \approx 4 K$
Boiling point of the solution = $85^{o} C$

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Similar questions

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If the boiling point elevation constant and the boiling point of benzene are 2.16 K kg mol−1 and 81.0∘ C respectively, calculate the boiling point of a solution formed when 5 g of C14H12 is dissolved in 15 g of benzene.

The correct option is D 85∘ CΔTb=kb×m molality= number of moles of soluteweight of solvent×1000 Molar mass of C14H12 = 180 g/mol molality=5/18015×1000=1.85 m ΔTb=kb×m =2.16×1.85≈4 K Boiling point of the solution = 85o C

If the boiling point elevation constant and the boiling point of benzene are $2.16 {K kg mol}^{- 1}$ and ${81.0}^{\circ} C$ respectively, calculate the boiling point of a solution formed when $5 g$ of $C_{14} H_{12}$ is dissolved in $15 g$ of benzene.