What is the final molarity of   KOH   in the   100. mL flask ?
817888
The number of moles of KOH used is:\n\n3.75 g / 56.1 g/mol = 0.0668 moles\n\nin 250 mL (1/4 L) this is 0.0688 * 4 = 0.267 M\n\nM1 * V1 = M2 * V2  or  M2 = (M1 * V1)/V2\n\n= (0.267 M * 27mL)/100mL = 7.209 / 100\n\n= 0.0721 M
564420
.1ml of 100:1 dilution had 280\n1.0 ml of dilution has 2800\n1.0 ml of undiluted has 280,000
747941
To provide 3.0 moles of KOH we need 2 L of a 1.5M KOH solution
422227
In your example, the concentration of the solution weight/weight would be (presuming the 100 ml is water, thus 100g) would be 10.7% (12 g solute/112 g total solution).
646163
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
837169
If you are asking what your words exactly are, then speaking in terms of an analytical chemist, you would be able to take only 3 aliquotes of exact 25 mL. After that, there is no doubt that 25 mL liquid is still in flask - there is no way by which you can squeez out the remaining 25 mL COMPLETELY from the flask. This is due to that a small amount , may be 1/2 an mL, would still be sticking to the walls of the flask. \nDont believe in me? Try it yourself !!!
19338
We will assume that 2.0 M NaCL(aq) is a dilute solution.  Therefore, 1 L of water weighs 1 kg.  For concentrated solutions, such as acids, you need to know the density of the solution.\n\nTo know the boiling point, use the formula\nDT = Kb * m\n\nWhere DT = change in temperature\nKb = boiling point constant (0.52 for water) \nm = molality.\n\nFirst, calculate the molality of the the solution\n\nMolality = moles of solute per kiograms of solven\n\n\n2.0 moles of NaCl / 1 kg solvent = 2.0\n\nDT=Kb * m\n= .52 * 2.0\n= 1.04 degrees\n= 1.0 degrees (using 2 significant digits)\n\nSo, water boils at 100 degrees C, so the boiling point of 2.0M NaCl is 101 degrees C.\n\nHope it helps.\n\n-Mark
482888
I'm guessing that your question pertains to the series of standard solutions you would prepare using the stock solution and volumetric flasks/pipette.  \n\nYes, you can use the C1V1=C2V2 formula, however there's a more practical and faster method that can be arrived at through common sense.  That is the concept of dilution factor.  If you have a stock solution of .5M iron for instance and you wanted to make a .05M iron solution, just dilute the next solution by a factor of 10%.  One way this can be achieved is by taking 10 mL of this stock solution using a volumetric pipette and adding it to a 100 mL volumetric flask.  To lessen the sampling error, it's better to take a larger sample volume, but usually this becomes tedious since lab equipment can be limited.\n\nSo take the original x molarity stock solution.  And if your final desired concentration is y, then the dilution factor is\n\ny/x \n\nThis percentage, y/x, represents the volume percentage in the final solution, that is multiply y/x by the particular To Deliver volume of the volumetric flask, to determine how much volume of the stock solution you need.......place this solution into the volumetric flask with a volumetric pipette, and then dilute to the mark with the standard solvent.
358982
NaOH is completely dissociated when it is dissolved in water.  A .0025M KOH solution therefore produces a [OH-] equal to .0025M.  The ion product of [H30+] and [OH-] is a constant, 1.0 x 10^-14 M^2.  By substitution, the [H30+] can be determined.  The pH can then be calculated.\n\n[H30+] = 1.0 x 10^-14 M^2/[OH-] = 1.0 x 10^-14 M^2/2.5 x 10^-3 \n\naaaand all that equals .4 x 10^-11... and you put that in correct scientific notation... 4 x 10^-12 M for the concentration of [H30+].\n\nWe then plug that in...   \n\npH= -log[H30+] = -log (4 x 10^-12) = 11.4\n\nThe pH of a .0025M solution of KOH is 11.4, which is greater than 7, making it a basic solution.\n\nStrong base at that.\n\nP.S.- If it is supposed to have the correct number of significant figures/digits, the answer would then be 11.
906606
1) To prepare this solution you should do:\n\n100 g of solution needs 2.5 g of solute (to get 2.5% mass/mass)\n150 g of solution needs  x    g of solute (to get 2.5% mass/mass)\n\n2)  x = (150 g x 2.5 g / 100 g) = 3.75 grames of KCl\n\nThat's it!\n\nGood luck!
736124
The answer to the first part:\n1) You need to find the pKa ---> use pKa+pKb = 14\nso 14-pKb = pKa\n14-4.75 = 9.25\n\n2) work out the final Molarity of each\n[NH3] = 0.400M/0.480L = 0.833M\n[NH4Cl] = 0.100M/0.520L = 0.192M\n\n3) now use the henderson-hasselbach equation\npH = pKa + log([base]/[acid])\n= 9.25 + log([0.833]/[0.192])\n= 9.89
1125846
NaVO3, MW = 122 g/mol\n\n100 mM = 0.1 M, so 0.1 moles in 1 L.  Dissolve 12.2g in water, dilute to 1L.
726441
4.74 / 300 = 0.0158 g/ml\nin 35 ml\n0.0158 * 35 = 0.553 g\nDividing by 40 g/mol (MW) we get\n0.553 / 40 = 0.0138 mol in 100 ml\nSo in 1000 ml the number of moles will be 0.138\nThe molarity is then 0.138 M.
501615
a 12.5% solution is 12.5 parts per 100.  To convert to parts per 1,000,000 you multiply by 10,000.\n\nSo 12.5% solution is 125,000 parts per million.\nTo dilute to 250 parts per million, you need a factor of \n125,000/250 dilution or 500.\n\nSo, depending on how much of the final solution you need, do a 500x dilution.  For example, I would use a volumetric pipet to transfer 1.00 mL of the 12.5% solution into a 500 mL volumetric flask, then dilute to the mark.
973272
ok, first go to moles.\n\n24.0g LiOH (1mol/23.948g) = 1.0022mol\n\nnow get liters using molarity\n\n1.0022mol (1L/0.153mol) = 6.55L or 6,550mL \n\nRight, there ya go, remember that you have 3 sig.figs. so you may want to represent significance by using scientific notation. So...\n6.55x10^3mL\n\nIf you're ever stuck, just simply go to moles and that will get you started (and usually gets you some points on tests).
850425
KOH + HNO3 ---> KNO3+ H20\n\nmoles of KOH = 0.025L*0.2mol/L = 0.005 mol\nconcentration of nitric acid solution = 0.005mol/0.030L = 0.167M\nconcentration of KNO3 = 0.005mol/0.055L = 0.091M
638399
The mole fraction of a component of a solution is simply the number of moles of the component divided by the total number of moles in the solution.\n\n\nmoles of C12H22O11 or solute = 29.6g(1mol/342.3g)=.086mol\n\n\nmole fraction= .086mol/x+.086=.193\n\nsolve for x,\n \nx=.360mol of solvent or water\n\n\nMolality= moles of solute/ kilograms of solvent.\n\nkilograms of solvent:\n \nmoles of water or solvent = .360\n\ngrams of water= 6.48\n\ncm^3 of water= 6.48\n\nkg of water = .0065\n\n Molality=m=.086mol of solute /.0065kg of solvent= 13.23m\n\nThe boiling point is:\n\n\n  13.23m*.512 degC/m= 6.77degC\n\n    100degC+ 6.77degC = 106.77degC