How do you convert osmolarity to osmolality?
414947
Osmolarity is in units of mol/L, while osmolality is in units of mol/kg. If the solvent for the solution is water, then there is no need for conversion, since 1 L of water weighs about 1 kg. So 0.5 molar solution of NaCl or any other substance will also be a 0.5 molal solution if the solvent is water.\n\nIf the solvent is not water (alcohol, benzene, etc) then you must convert the mol/L into mol/kg. The easiest way to do this is to find out what the solvent's density is.\n\nFor example: imagine you have a 0.500 molar solution of napthalene dissolved in benzene. That means that there is 0.500 moles of napthalene for every 1 liter of benzene (since we are talking about molarity). The density of benzene is 0.8765 g/ml, which is equivalent to 0.8675 kg/L. So use the density to help you solve for the molality:\n\nMOLARITY / DENSITY = MOLALITY\n\nIn our example:\n\n0.500 mol napthalene/L       \n------------------\n0.8675 kg/L\n\nThe liters cancel out and you are left with the units of mol/kg (molality). In this case, the answer is 0.576 mol/kg.
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.
710157
ANSWER: 0.482 M\n\n\nEXPLANATION:\n\nM = mol solute / L soln\n\nTherefore, you need to find the mol of solute and L of soln to find M.  \n\nTo find mol solute:\n10800mg Na(1g Na/1000mg Na)(1mol Na/22.99g Na) = 0.4698mol Na\n\nTo find L of soln:\n1kg soln(1000g soln/1kg soln)(1ml soln/1.025g soln)(1L soln/1000ml soln) = 0.9756L\n\nUsing the two values found plug in to M = mol solute / L soln\n\n0.4698mol Na / 0.9756L soln = 0.482 M
601420
Well, I'm not doing your homework for you, but Molarity is Moles per liter.  So that means you would have to change grams to moles, hopefully you know how to do that.
628177
HCL   +   H2O     ->   H2O     +    H+    +   CL-\n1X10-4 M  1X10-4M  ->  1X10-4M    1X10-4M     1X10-4M\nPh = -log (1x10-4)\n   = 4\nPoh=14-4\n   =10\n(OH-) conc. in Mol/dm3 = 1x10-10 M
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
944773
The dilution formula is:\n\n(M1 and V1 are the concentration and volume of the stock solution; M2 and V2 are the concentration and volume of final solution.  M1 and M2 should be in molarities; V1 and V2 should be liters, but it doesn't matter as long as they're the same units)\nM1V1 = M2V2\n(12.0 M)(V1) = (2.50 M)(250.0 mL)\nV1 = 52.1 mL\n\n(C)
639379
The boiling point elevation constant of water is required\n\nThe formula is T = Kfm where m is molality and T is rise in temp.
586791
MOLARITY=(MOLECULAR WT/GIVEN WT)*(1000/VOLUME OF THE SOLUTION IN MILLIITERS)\n\nBUT WE KNOW THT M1V1=M2V2\nie 35*2.104=M2*825\nM2=0.0896M\nTHE AMOUNT OF WATER ADDED IS 825-35=790.
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
955628
For the reaction:\n\nMgO + 2 HCl  -->  MgCl2 + H2O\n\nThe temperature should go up, not down.  Only the initial and final temperature are needed.\n\nTypical values for students are between 110 an 150 kJ/mol MgO.  This is typical, not the best that can be done.
371216
Molarity is #moles/ volume in liters.\n\nFirst calculate the # of moles NaOH using the molecular weight (MW=40g/mole) of NaOH.\n\n#moles NaOH = weight NaOH/MW = 100g/(40g/mol)= 2.5moles\n\nNow calculate the molarity \n\nM= #moles NaOH/ volume = 2.5moles/20L = 0.125 Molar
1154934
Online conversion figures these things out for you!\n\nYou have to choose liquid or dry matter.  But the answer is there.
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
807846
Molarity is mol per litre of solution.\n\nAssuming the volume of salt is negligible, we just add 1 mol of NaCl to 1 litre of water.\n\nTo find 1 mol of NaCl, add the molar mass of each substance.\n22.99 +35.45 = 58.44g\n\nSo, you add 58.44g of salt to 1 litre of water.\n\n(If you want to take into account the volume of salt, find the density to determine the volume, and subtract that volume from 1L of water to determine the amount of water to add, though generally we just assume the volume is negligible)\n\nNOTE: This is only a 1 molar solution for the temperature and presure at which you made it. The problem with molarity is that it is mol per litre of solvent, and the volume of a substance can vary with changes in temperature and pressure.
793017
Take 1.875 L of the .2 molar stock solution of NaBr and add .625 L of water to it and you get 2.5 L of a .15 molar solution of NaBr this just use the c1v1=c2v2 equation.\n\nNote:\nThis is the only way it could work. You do not want to add more NaBr because your concentration is already higher then desired.   If you just added 0.5L of water to the 2.0 L of .2 molar stock solution your molarity would still be to high at .16 meaning you have to dilute a smaller portion of your 2L .2 molar stock solution.