Dut: MyC × .. 4.) SolutionproblemSet,Ke x / Solutions Math Homevc Secure https/g
ID: 196254 • Letter: D
Question
Dut: MyC × .. 4.) SolutionproblemSet,Ke x / Solutions Math Homevc Secure https/georgiasouthern.desire2learn.com/d2/le/content/372757/fullscreen/6085695/View Math Homework you have the following substances: A bottle of stock 70% Ethanol (liquid: mw=46.07) A bottle of stock 4 M PBS (liquid) A bottle of stock TBS (liquid) A bottle of stock 1M NaHCO; (liquid:mw 84.07) A bottle of sucrose crystals (solid: mw 342) A large bottle of NaC1 (solid: mw-58) A large bottle of agarose (solid: mw=306) 1 bottle of NaH2PO4 (solid: mw 120) 1 bottle ofNa2HP04 (solid: mw=142) Bottle of estradiol (solid; mw 272.38) Huge jug of TAE (liquid) Huge jug of distilled water For each solution, clearly indicate the amount of EACH solute and amount of solvent you would us prepare the solution. Identify the solvent and solute by name and indicate the amount using proper u (gms or mls), unless otherwise indicated. Assume water is the solvent if not indicated 1, 850 mls of 3% TBS buffer 2.60 mls of 1.5% agarose in TAE 3. 1000 mls of 0.25 M NaCl 112Explanation / Answer
1. 850 ml of 3% TBS buffer: It can be prepared by adding 25.5 ml of stock TBS liquid into 824.5 ml of water. 850 ml of 3% TBS buffer means there will be 3% TBS in total 850 ml of TBS buffer solution. 3% of 850 ml = 25.5 ml. Therefore by talking 25.5 ml of stock TBS liquid and make up the volume upto 850 ml with water (total water = 850-25.5 =824.5 ml) will give 850 ml of 3% TBS buffer solution.
2. 60 ml of 1.5% agarose in TAE. It can be prepared by taking 0.9 gm of agarose in a flask and make up the volume upto 60 ml with TAE. 1.5% agarose in TAE indicates 1.5 gm agarose in 100 ml of TAE. Since 100 ml required 1.5 gm of agarose therefore 60 ml require 1.5 X 60/100 = 0.9 gm. Thus, by taking 0.9 gm of agarose and making up the volume upto 60 ml with TAE will give 60 ml 1.5% agarose in TAE.
3. 100 ml of 0.25M NaCl: It can be prepared by taking 1.45 gm of NaCl into 100 ml of water. Since 1M NaCl solution indicates 58 gm NaCl in 1000 ml of water therefore 0.25M NaCl will be 14.5 gm NaCl in 1000 ml water. Thus, 1.45 gm of NaCl in 100 ml of water will give 100 ml of 0.25M NaCl solution.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.