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Exercise
Practical 8 - Estimation of Sugars (Reducing Sugar) |
Aim: Estimation of reducing sugars in fresh fruits and processed products using Lane and Eynon method.
Theory: Sugars are inherently present in different fruits and vegetables or added during preparation of different products. They are added to the fruit products to improve the taste and also to act as a preservative. Glucose and fructose in the fruits represent reducing sugars while sucrose or cane sugar added represents the non-reducing sugar. They are estimated by using Lane and Eynon method which measures sugar as reducing sugar and total sugar as invert sugar. Principle: Invert sugar reduces the copper in Fehling’s solution to red, insoluble cuprous oxide. The sugar content in a food sample is estimated by determining the volume of the unknown sugar solution required to completely reduce a known volume of Fehling’s solution. Glucose and other sugars are capable of reducing oxidizing agents and are called reducing sugars and this property is used for the estimation of sugars. The cupric ion in Fehling’s solution is reduced to cuprous state which precipitates as red cuprous oxide (Cu2O). Only reducing sugars reduce the copper solution. The method is suitable for estimation of sugars in fruit and fruit products. CuSO4 + 2NaOH → Cu (OH)2 + Na2SO4 Cu (OH)2 → CuO (Cupric oxide) 2CuO + CHO → Cu2O (Cuprous oxide) + COOH Apparatus, reagents and glassware required • Beakers - 250ml • Volumetric flasks - 250ml • Measuring cylinder- 250ml • Pipette - 10ml • Hot plate Preparation of Reagents
Standardization of standard invert sugar: Pipette 25ml of standard invert solution in to a 100ml volumetric flask; add 50ml water and few drops of Phenolphthalein indicator. Neutralize with 20% NaOH until solution turns pink. Acidify with 1N HCl adding it drop wise until pink color disappears. Make up to 100ml with water (1ml=2.5mg invert sugar). Standardization of Fehling’s solution: Mix 5ml Fehling A + 5ml Fehling B solution in 250ml conical flask. Add 25-50ml water and heat the flask. Add standard invert sugar solution from the burette dropwise till the solution turns brick red. Add few drops of Methylene blue indicator and add drop-wise invert solution, when the blue color disappears, note the titre value of invert solution, repeat the titration and calculate factor for Fehling’s solution as under:- Titre x 2.5 Factor for Fehling solution((g of invert sugar) ) = --------------------
1000 Procedure for estimation of reducing sugars Take 10-20g juice/squash/drink in 250 ml volumetric flask, add 100 ml water neutralize with 1 N NaOH. Add 2ml 45% lead acetate. Shake well and keep for 10 minutes. Add few drops of potassium oxalate solution to remove excess of lead acetate. Make volume to 250ml with water and filter. OR Take 10-25 g of sample (fresh fruit or fruit product) and grind in a pestle and mortar, blend in blender, add 100 ml water. Neutralize solution with 1 N NaOH. Boil gently for 1 hour with stirring. Replace water lost during evaporation, cool and transfer to 500 ml volumetric flask. Make volume 500 ml and filter through whatman filter paper.
Calculations mg of invert sugar × dilution × 100 Reducing sugar (%) = --------- --------------------------------------- titre × weight of the sample × 100 Or
Factor for Fehling solution × Volume made × 100 % Reducing Sugar = -------------------------------------------------------- Titre × wt of sample × ml of aliquot Problem: 10 g of fruit jam was made into 500 ml after neutralization with 1 N NaOH and filtered. From the filtrate 100ml of aliquot was clarified using lead acetate and potassium oxalate and volume made to 250 ml and filtered. 10 ml Fehling’s solution (A+B) was titrate against clarified sample. The titre value was 20 ml. Calculate the reducing sugar when factor for Fehling’s solution is 0.05. Solution:
Factor× vol. made for 10g to 500ml × vol made100ml to 250 ml × 100 % Reducing Sugar = -------------------------------------------------------- Titre × wt of sample (10) × ml of aliquot 100 0.05 × 500 × 250 × 100
= ---------------------------------
20 × 10 × 100 = 31.25%
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Last modified: Wednesday, 7 March 2012, 9:59 AM