You need to prepare 250. mL of a 0.200 M aqueous solution of sucrose?
You need to prepare 250. mL of a 0.200 M aqueous solution of sucrose, C12H22O11 (aq), which is used frequently in biological experiments. Based on your answer above, what is the value of x? A 0.200 M aqueous solution contains 0.200 mole of sucrose per liter of solution. 250 ml = 0.250 liter 250. mL of a 0.200 M aqueous solution of sucrose contains 0.250 * 0.200 = 0.050 mole of sucrose Mass = number of moles * mass of 1 mole mass of 1 mole of C12H22O11 = 144 + 22 + 176 = 342 grams Mass = 0.05 * 342 = 17.1 grams of sucrose. Dissolve 17.1 grams of sucrose in enough water to produce 250 ml of solution. I dissolve the sucrose in 200 ml of water, and fill to 250 ml.
Calculating the pH of a buffer solution?
Hi, well this isn't so bad I would be glad to help. First off we need to decide which of these is our acid and which is the base. HCHO2 is formic acid - so it's the acid NaCHO2 is sodium formate. In an aqueous solution it is present as a sodium ion (Na+) and a formate ion (CHO2-) because of the negative charge on the formate it can accept an H+ and is a base. Now since we're working with a buffer we can use the Henderson–Hasselbalch equation: pH = pKa + log (base/acid) Quick note - in this equation you can use either moles of base and acid or concentration of base and acid, but you can only use concentration if the base and acid are contained in the same volume. Since the base and acid are in different volumes here we have to solve for the number of moles of each. To find the moles of base (sodium formate) we do this: 15 mL x (1 L / 1000 mL) x (0.5 mol / 1 L) conversion to L molarity of base When you solve you get : 0.0075 mol sodium formate Do the same for acid (formic acid): 60mL x (1L / 1000 mL) x (0.250 mol / 1 L) And you get: 0.015 mol formic acid Now we have all of the parts of the equation we need except pH (which we're solving for) and pKa. We HAVE to have the pKa and luckily for us it's a physical constant so you can look it up anywhere. Happens to be 3.744. Let's start plugging information into the equation: pH = 3.744 + log (0.0075/0.015) On your calculator make sure you do the log first if you have a scientific calculator or enter it like this 3.744+(log(0.0075/0.15)) if you have a graphing calculator. If you enter it correctly you get pH = 3.44! Hope this helped!
What volume of concentrated H2SO4 (density = 1.84g/mL and 96% purity) would be required to prepare 500mL of a 0.2000M H2SO4 solution?
Dear anonymous,Apparently, you want us to do your homework for you. Below is the solution to your question. Do know that this solution is useless to you unless you understand each step and why it is taken. So while I give you the solution to the question, it will be up to you to study this answer carefully and make sure you understand it.To start of with, we will need to fix the question itself as it contains a rather big error. The density of any substance is given in g/cm^3. Not in g/mol as that would be the molecular weight. So the density is 1.84 g/cm^3 (=1.84 g/mL).The answer:0.2M = 0.2 mole/LTherefore, 0.5L * 0.2M = 0.1 mole of H2SO4 required.The molar mass of H2SO4 is 98.079 g/mole (see periodic table or google).Therefore, 98.079 g/mole * 0.1 mole = 9.8079 g9.8079 g/ 1.84 g/cm^3 = 5.33038 cm^3 (=5.33038 mL)The purity of your substance is 96%, therefore you will actually need:5.33038*(1/0.96) = 5.55248 mL.The required precision in your answer is 3 digits (due to the precision of 500 mL).Therefore, the answer is 5.55 mL of acid and (500–5.55 =) 494 mL of water.Note that when preparing such a dilution, always add acid to water. Never the other way around. Pure H2SO4 is an extremely dangerous substance. Handle only under the supervision of a professional.
Chem help please? solutions and molarity?
You're going to use MV = grams / molar mass for lots (but not all) of these calculations. Note that the volume must be in liters when using this equation. 1) (M) (0.500 L) = 25.0 g / 294.2 g/mol M = 0.170 mol/L Use this: M1V1 = M2V2 for the second part of question #1. The volumes do not have to be in liters, but they must be in the same units (mL in the case just below): (0.170 mol/L) (15.0 mL) = (x) (100.0 mL) x = 0.0255 mol/L 2. A solution of ethanol (d=0.795 g/mL) and water (H2O) is prepared by placing 30.0 mL of ethanol in a 250.0 mL volumetric flask and diluting with water to the flask etched mark. 30.0 mL times 0.795 g/mol = 23.85 g of ethanol use MV = grams / molar mass (M) (0.250 L) = 23.85 g / ____ At ___, put the molar mass of C2H5OH. For part b, use M1V1 = M2V2 Problem 3: M1V1 = M2V2 for part a. You'll need to molar mass of Na2CO3 for part b. Problem 4 is like 3, just with Na3PO4 for part b. Solutions stuff: http://www.chemteam.info/Solutions/Solut... "please, serious answers only, this is for some homework that i am totally lost on and really need to do well on. thank you" 1) I wish you well in this HW and all others. 2) You are quite welcome. I hope I have helped.
Determine the volume (mL) required to prepare each of the following.?
A. 10.0 mL of a 0.300 M KNO3 solution from a 7.50 M KNO3 solution. B. 10.0 mL of 4.00 M H2SO4 solution using a 11.0 M H2SO4 solution. C. 0.200 L of a 1.00 M NH4Cl solution using a 11.5 M NH4Cl solution.