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ChemistryLiquid SolutionColligative PropertiesMedium2 minQB

ChemistryMediumnumerical

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 is the average molar mass (in gm/mol) of the dissolved materials ?

Answer:

217.00

Solution:

Molality ( $m$ ): $Δ T_{b} = 100.27℃ - 100℃ = 0.27 K ⟹ m = \frac{Δ T _{b}}{K _{b}} = \frac{0.27}{0.512} = 0.5273 m o l / k g$

Degree of Inversion ( $α$ ): $m = \frac{0.02924 ( 1 + α )}{( 90 - 0.5263 α ) \times 1 0 ^{- 3}} = 0.5273 ⟹ α \approx 0.617$

Total Mass of Dissolved Materials: $W = 10 g + (18 \times 0.02924 \times 0.617) g = 10.325 g$

Total Moles of Dissolved Particles: $n = 0.02924 \times (1 + 0.617) = 0.04729 m o l$

Average Molar Mass ( $M_{a vg}$ ): $M_{a vg} = \frac{W}{n} = \frac{10.325}{0.04729} = 218.33g/mol$

Stream:JEESubject:ChemistryTopic:Liquid SolutionSubtopic:Colligative Properties

⏱ 2mℹ️ Source: QB

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