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Standard XII

Chemistry

Characteristics of Chemical Equilibrium

The optical r...

Question

The optical rotation of the $α - f o r m$ of a pyramose is $+ {150.7}^{\circ}$ , that of the $β - f o r m$ is $+ {52.8}^{\circ}$ . In solution an equilibrium mixture of these anomers has an optical rotation of $+ {80.2}^{\circ}$ . The percentage of the $α$ form in equilibrium mixture is:

28

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Solution

The correct option is A $28$ %
$α$ from $= + 150. 7^{0}$ , $β$ from $= + 52. 8^{0}$
at equilibrium optical rotation $= + 80.2$
let, at equilibrium $α$ -from exist $= x$
at equilibrium $β$ -from exist $= (100 - x)$
Therefore, $\frac{150.7 x + (100 - x) \times 52.8}{100} = 80.2$
$\Rightarrow 150.7 x + 5280 - 52.8 x = 8020$
$\Rightarrow 99.9 x = 2740$
$\Rightarrow x = 27.42 \approx 28$ %

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The optical rotation of the α−form of a pyramose is +150.7∘, that of the β−form is +52.8∘. In solution an equilibrium mixture of these anomers has an optical rotation of +80.2∘. The percentage of the α form in equilibrium mixture is:

The correct option is A 28%α from =+150.70, β from =+52.80at equilibrium optical rotation =+80.2 let, at equilibrium α-from exist =x at equilibrium β-from exist =(100−x)Therefore, 150.7x+(100−x)×52.8100=80.2⇒150.7x+5280−52.8x=8020⇒99.9x=2740⇒x=27.42≈28%

The optical rotation of the $α - f o r m$ of a pyramose is $+ {150.7}^{\circ}$ , that of the $β - f o r m$ is $+ {52.8}^{\circ}$ . In solution an equilibrium mixture of these anomers has an optical rotation of $+ {80.2}^{\circ}$ . The percentage of the $α$ form in equilibrium mixture is: