The correct option is A 0.2Let degree of dissociation of sucrose be #160; α ⇒ #160;Sucrose + water→ glucose + fructose The molecular mass of sucro ...

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Standard XII

Chemistry

Polarisability and Fajan's Rule

A 10

Question

A 10% (w/w) solution of cane sugar has undergone partial inversion according to the reaction:
sucrose + water $\to$ glucose + fructose
The boiling point of the solution is ${100.7}^{o} C$ . What fraction of sugar has inverted? $[K_{b} (H_{2} O) = 1.8 K g K m o l^{- 1}]$

0.5

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0.2

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Solution

The correct option is A 0.2
Let degree of dissociation of sucrose be $α$
$\Rightarrow S u c r o s e + w a t e r \to g l u c o s e + f r u c t o s e$
The molecular mass of sucrose is 342.
$i = 1 + (2 - 1) α = 1 + α$
$\Rightarrow Δ T_{b} = i K_{b} m \Rightarrow 0.7 = (1 + α) \times 1.8 \times \frac{10 \times 1000}{342 \times 90}$
$\Rightarrow 1 + α = \frac{0.7 \times 342 \times 9}{1.8 \times 1000} = 1.2$
$\Rightarrow α = 0.2$

Option B is correct.

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Similar questions

Q. A

10

per (w/w) solution of cane sugar has undergone partial inversion according to the reaction:

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

If the boiling point of solution is

{100.27}^{\circ} C

(a) what is the average mass of the dissolved materials?

(b) what fraction of the sugar has inverted?

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

Q. 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

0.512 K m o l^{- 1} k g)

Multiply answer with 10 and write the nearest integer value.

Q. A

5

% solution (w/W) of cane sugar (molar mass

= 342 g

{m o l}^{- 1}

) has freezing point

271 K

. What will be the freezing point of

5

% glucose (molar mass

= 18 g

{m o l}^{- 1}

) in water if freezing point of pure water is

273.15 K

Q. Cane sugar underoges the inversion as follow:

C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}

If solution of 0.025 moles of sugar in 200 gm of water show depression in freezing point 0.372

^{o}

C, then what % sucrose has inverted.

(K_{f} (H_{2} O) = 1.86 K k g m o l^{1})

Q. A

5

% solution (by mass) of cane sugar in water has freezing point of

271 K

and freezing point of pure water is

273.15 K

. The freezing point of a

5

% solution (by mass) of glucose in water is :

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A 10% (w/w) solution of cane sugar has undergone partial inversion according to the reaction:sucrose + water → glucose + fructoseThe boiling point of the solution is 100.7oC. What fraction of sugar has inverted? [Kb(H2O)=1.8 KgKmol−1]

The correct option is A 0.2Let degree of dissociation of sucrose be α⇒Sucrose+water→glucose+fructoseThe molecular mass of sucrose is 342.i=1+(2−1)α=1+α⇒ΔTb=iKbm⇒0.7=(1+α)×1.8×10×1000342×90 ⇒1+α=0.7×342×91.8×1000=1.2 ⇒α=0.2Option B is correct.

A 10% (w/w) solution of cane sugar has undergone partial inversion according to the reaction:
sucrose + water $\to$ glucose + fructose
The boiling point of the solution is ${100.7}^{o} C$ . What fraction of sugar has inverted? $[K_{b} (H_{2} O) = 1.8 K g K m o l^{- 1}]$