The elevation in the boiling point #160; Δ T_b = 100.27 100 = 0.27 ^oC #160; Δ T_b =K_b m #160; 0.27 = 0.512 × m Molality m = 0.527 ...

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Colligative Properties

A 10

Question

A 10% (mass/mass) solution of cane sugar undergoes partial conversion into glucose and fructose to show an inversion of cane sugar as:
$S u c r o s e + W a t e r \to G l u c o s e + F r u c t o s e$

The solution boils at ${100.27}^{o} C$ at this state. What fraction of the sugar has inverted? (Given $K_{b}$ for $H_{2} O$ is $0.512 K m o l^{- 1} k g)$
Multiply answer with 10 and write the nearest integer value.

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Solution

The elevation in the boiling point
$Δ T_{b} = 100.27 - 100 = {0.27}^{o} C$
$Δ T_{b} = K_{b} m$
$0.27 = 0.512 \times m$
Molality $m = 0.527$ m
$10$ % solution means $1000$ g (1 kg ) of water has $1000 \times \frac{10}{90} = 111.1 g$ of sucrose present before inversion.
Molar mass of sucrose is $342$ g/mol. Number of moles of sucrose present intially $= \frac{111.1}{342} = 0.3248$ moles.
After inversion, the molality is $0.527$ m. So $1$ kg of water will contain $0.527$ moles of solute.
The fraction of the sugar has inverted $= \frac{0.527 - 0.3248}{0.3248} = 0.623$

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10

S u c r o s e + W a t e r \to G l u c o s e + F r u c t o s e

{100.27}^{\circ} C

K_{b} (H_{2} O) = 0.512 K m o l^{- 1} k g

\to

{100.7}^{o} C

[K_{b} (H_{2} O) = 1.8 K g K m o l^{- 1}]

An enzyme that converts cane sugar into glucose and fructose is:

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