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ChemistryLiquid SolutionColligative PropertiesMedium2 minQB

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Passage / Comprehension

A 10% w/w solution of cane sugar (Sucrose, $C_{12} H_{22} O_{11}$ ) has undergone partial inversion according to the reaction: Sucrose + Water → Glucose + Fructose. If the boiling point of solution is 100.27°C. $K_{b} (H_{2} O) = 0.512 K mol^{- 1} kg$

What percentage of the sugar has inverted?

Answer:

61.00

Solution:

For 10% w/w solution:

Sucrose = 10 g, Water = 90 g = 0.09 kg

$Δ T_{b} = 100.27 - 100 = 0.27 K$

$Δ T_{b} = i K_{b} m$

Initial moles of sucrose:

$n = \frac{10}{342} = 0.0292$

$m = \frac{0.0292}{0.09} = 0.324$

$0.27 = i (0.512) (0.324)$

$i = \frac{0.27}{0.166} = 1.62$

For partial inversion:

If $x$ mole sucrose inverts, particles increase by $x$ .

$i = \frac{1 + x}{1} = 1.62$

$x = 0.62$

Therefore, the percentage of sugar inverted = 62%.

Stream:JEESubject:ChemistryTopic:Liquid SolutionSubtopic:Colligative Properties

⏱ 2mℹ️ Source: QB

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