The specific rotation of the equilibrium mixture is +52.7 deg mL-1 g-1 dm-1 . Th

Question

The specific rotation of the equilibrium mixture is +52.7 deg mL-1 g-1 dm-1 . The specific rotation of pure -D-glucose is +18.7 deg mL-1 g-1 dm-1 and that of pure -D-glucose is +112 deg mL-1 g-1 dm-1. Calculate the percentage of each form of glucose present in the equilibrium mixture. Show your work below.

Percentage of -D-glucose: _____________________

Percentage of -D-glucose: _____________________

The specific rotation of the equilibrium mixture is +52.7 deg mL-1 g-1 dm-1 . The specific rotation of pure I^2-D-glucose is +18.7 deg mL-1 g-1 dm-1 and that of pure I + - D-glucose is +112 deg mL-1 g-1 dm-1. Calculate the percentage of each form of glucose present in the equilibrium mixture. Show your work below. Percentage of I^2-D-glucose: _____________________ Percentage of I^2 + -D-glucose: _____________________

Explanation / Answer

Solution :-

Given data

specific rotation of the equilibrium mixture is +52.7 deg mL-1 g-1 dm-1

specific rotation of pure -D-glucose is +18.7 deg mL-1 g-1 dm-1

specific rotation of pure -D-glucose is +112 deg mL-1 g-1 dm-1

percentage of each form of glucose in equilibrium mixture.

Lets assume fraction of the pure -D-glucose = x

Then fraction of the -D-glucose is in the mixture = 1-x

Now lets make set up to calculate the fraction of the each glucose in the equilibrium mixture.

52.7 = (112 *x ) + (18.7 * (1-x))

52.7 = 112x + 18.7 – 18.7 x

52.7 – 18.7 = 112x – 18.7 x

34 = 93.3x

34/93.3 = x

0.3644 = x

So the fraction of the -D-glucose = 0.3644

And fraction of the -D-glucose is = 1-x = 1-0.3644 = 0.6356

Now lets convert these fractions into percentages

Percentage of the -D-glucose = 0.3644 * 100% = 36.44 %

Percentage of the -D-glucose = 0.6356 * 100 % = 63.56 %

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