Question

How do you make a 10, 20, and 30% solution of sucrose?

   b. A student measured out 0.509g of galactose and dissolved it in 25.0mL of water. 10.0 cm of sample was added to the cell, producing an angle of 204.4. If the blank's angle was 202.7, what is the observed angle of rotation for the sugar solution, the specific rotation of the sugar, and the percent difference.

c. Calculate the concentration of sucrose using the following information:  blank's angle = 180.0, sample height = 9.7cm, and sample's angle = 186.3.

Answer 1

Sol :-

(a). 10 % solution of sucrose (weight/volume) means that 10 g of sucrose dissolved in 100 mL of water. Also, 20 % solution of sucrose means 20 g of sucrose dissolved in 100 mL of water. 30 % solution of sucrose means that 30 g of sucrose dissolved in 100 mL of water.

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(b). Given, Mass of galactose = 0.509 g

angle = 202.7⁰

Volume (V) = 25 mL

Path length (l) = 10 cm

Angle of rotation of galactose solution = 204.4 ⁰

Angle of rotation of galactose molecule (angle) = 204.4⁰ - 202.7⁰ = 1.7⁰

Now,

Specific rotation = Angle x Volume (V) / (Mass x Path length )

= (1.7⁰ x 25 mL ) / (0.509 g x 1.0 dm)

= 83.5 ⁰.ml.g^-1.dm^-1

Also, specific rotation in standard state = 81⁰

Percent error = (83.5⁰ - 81⁰) / 81⁰ x 100 %

= 3.09 %

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(c).

Again , l = 9.7 cm

angle = 186.3⁰

Specific rotation = 66.37 ml.g^-1dm^-1

banck's angle = 180⁰

So,

Angle of rotation = 186.3⁰ - 180⁰

= 6.3⁰

and

Specific rotation = Angle / Concentration x Path length

Concentration = 6.3⁰ / (66.37 ml.g^-1dm^-1 x 0.97 dm)

Concentration = 0.098 g/mL