1. You fill each of three (10 ml) dialysis tubes with one of the following subst
ID: 210234 • Letter: 1
Question
1. You fill each of three (10 ml) dialysis tubes with one of the following substances: a. 2% by weight of sucrose (MW 342.297) in RO water. b, 2% by weight bovine serum albumin (MW 66,000) in RO water. The dialysis tubing is impermeable to the solutes. Each bag initially weighs 10.0 g. Bags are placed into separate beakers of RO water. After two hours the bags are removed and the weight recorded. How do you explain the difference in final weights of the dialysis tubes in the table below? Dialysis Tube Weight 13.2g Bovine Serum Albumin 10.9 g Glucose 2. How much 0.20 M sucrose can be made from 250 ml of 0.5 M sucrose? A patient may be diagnosed with diabetes if he/she has a blood glucose level great than 200 mg di what is the molarity (in ml) of this glucose level? (MW = 180.156) (dl = deciliter = 1/10 of a liter). 3.Explanation / Answer
Answer
Osmosis: The movement of water from its higher concentration to its lower concentration through semi-permeable or selectively permeable membrane.
When dialysis tube is filled with these solution one is 2% sucrose solution and 2% bovine serum albumin (protein) and both the tube separately place in the RO water where they have higher concentration of water out side to the tube
In the tube contain sucrose solution highly increase the osmolarity of the solution but tube contain bovine serum albumin (protein) is little increase the osmolarity.
So, in the experiment tube with sucrose increases the weight i.e. 13.2g as compare to the initial weight 10g. But protein contain solution only little increase the weight i.e. 10.9
When we calculate the cumulative percent weight change
Then change = final weight - initial weight / initial weight*100
Sucrose bag change = 13.2- 10/ 10*100
= 32%
Protein bag change = 10.9- 10/ 10 * 100
= 9%
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.