Question

You have a 5.0 ml sample of a protein in 0.5M NaCI. You place the protein/salt sample inside dialysis tubing and place the bag in a large beaker of distilled water. Your goal is to remove as much NaCI from the sample as possible. Since you know that the final salt concentration will be the same inside the tubing as out, you can calculate the final concentration of the salt. If you decide to place the dialysis bag in 4.0 L of distilled water for 12 h the final concentration of the salt would be mM. the tolerance is +/-5% What would the final salt concentration be if instead of placing the dialysis bag in 4.0 L of distilled water for 12 h, you placed the bag in 1.0 L of distilled water for 6.0 h and then in another 1.0 L of fresh distilled water for another 6.0 h? It would be mM.

Answer 1

sample 0.5 M Nack 0.005L) 5 mL 00025 moles oNac 0.5 mole X Nac=58 44g/mo (0.025 X58-44) 0-4618 dilution concenbnation is Aame Lets say »q. in dialy sis bag and (0-1461-38 Nac outside fter O 1461-moles 58 A4 58.44 0.005 L NaCl coneentration 4 L ( 0.00018 mo 42= 0.005 0-1461-) 58.44 4.005 0.00073 - 0.000189 0.005 L 0.62 mM Ip ILfinut dialysis-> 0-1461- 58.44 58.44 -00 51 0.00 073 > 0.000726 0 DO 5

Coneemt ation 0.000726 2-48 mM 58-44 0-005 Second dilutiorn- 0.0024 8 M 24 Xiô5moles x 0-005L Na c 0000726 8 amount ofter dilution Ygm Nack in si cde and 0 000726g outside. 0000726-y 58.4 4 58.44 0.005 -005 3.63x156 Concertration Y3.61X0 3. 6XIO 58.44 0.012 mM Final 0. 0 o5 12- 35X10 M O.012 mM