12) You are asked to prepare125.0 mL of 0.0321 M AgNO3. How many grams would you
ID: 685526 • Letter: 1
Question
12) You are asked to prepare125.0 mL of 0.0321 M AgNO3. How many grams would you need of a
14) Blood cholesterol levelsare generally expressed as milligrams of cholesterol per deciliterof
blood. What is theapproximate mass percent cholesterol in a blood sample having acholesterol
4) Some vitamins are watersoluble and some are fat soluble (Fats are substances whosemolecules have
long hydrocarbon chains).The structural formulas of two vitamins are shown in thetext–one is water
soluble and one is fatsoluble. Identify which is which, and explain yourreasoning.
(a) VitaminC:
(b) Vitamin E:
Explanation / Answer
12) (Please rate each individual response as you are askingthree different questions) I hope I understand yourquestion correctly. This is a multiple step problem. The key is tofirst calculate the moles in the desiredsolution. Then calculatethe weight of the provided sample necessary to yield that amount ofmoles. M = molarity = moles solute / liters of solution MMAgNO3 = 169.87 g/mol Solution = solute + solvent You are given 125.0 mL of 0.0321 M AgNO3, you can calculate the moles in that solution by using thegiven volume of solution. ( volume (L) of solution molarity > moles of solute) (125.0 mL soln)(1 L soln/1000mL soln)(0.0321 moles AgNO3 / 1 Lsoln) = 0.004013 molAgNO3 This is the amount of moles necessary to prepare thesolution. You can then calculate how much you need toweigh out from a 99.81% by mass sample of AgNO3. Assume you have 100 g of the sample, then there is 99.81 gof AgNO3 in that sample. (100 g sample / 99.81 g AgNO3)(169.87 gAgNO3 / 1 mol AgNO3)(0.004013 molAgNO3) = 0.683 g of 99.81%sample This means if you weigh out 0.683 g ofthe 99.81% AgNO3 sample, then dissolve it in 125.0 mLH2O, it will yield the desired 0.00401 mAgNO3 soln
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