Question

The payment sugar beet growers receive is based on the sucrose content of their delivered beets. A sample is taken from each load of beets and the sugar content in the beets is determined by polarimetry of the beet extract. In a typical solution 50 grams of pulverized beet tissue is homogenized with 360 mL of dilute aluminum sulfate. The homogenate is filtered and the optical rotation is determined by use of a polarimeter. For one sample, the following observations were noted:

Tube length: 4 decimeters

Temperature: 20 degrees Celsius

Optical rotation (a) = +4.97 degrees

Total volume of extract= 405 mL

[a]D = a(100)/(1)(c)

[a]D= specific rotation of light of wavelength, at t °C (from table below). l = length of polarimeter tube in decimeters. c = concentration of sugar in g/l00 ml of solution. a = observed rotation in angular degrees (polarimeter measurement).

1. How much sucrose was contained in the 50 gram sample?

2. If the sugar beet grower receives $0.145 for each pound of sugar, what would he receive for one metric ton (1000 Kg) of beets?

Answer 1

since [a]_D = a(100)/(l)(c)

[a]_D = 66.5 degrees for sucrose

a) hence according to the given values,  66.5 = 4.97/(4*c), which means c = 0.0187g/100mL

hence sucrose contained in 405mL sample = (0.0187*4.05) = 0.0757g i.e 50g sample of beet root contains 0.0757 g sucrose

b) sucrose extracted from 50g beets = 0.0757g

sucrose extracted from 1kg beets = 0.0757g*20 = 1.514g

sucrose extracted from 1000kg beets = 1.514g*1000 = 1514g

1 gram = 0.00220462 pounds

1514 grams = 0.00220462*1514 = 3.33779468 pounds

money received for 1 pound = $0.145

money received for 3.33779468 pounds = 3.33779468*0.145 = $0.484