would the calculated molar mass of the solute be too high or too low?
124024
If the freezing point of the solution had been incorrectly read .3̊ lower than the true freezing point, then the observed ΔTf would have been greater than it should have been. This would have caused the calculated molality to be too high. If the calculated molality were too high, then the number of moles present as calculated would be too high. An\nexaggerated number of moles would cause the calculated molar mass to be too low, as the same weight of the substance would be distributed against more moles of that substance.\nThe substance would seem lighter than it is.\n\nI'm sure this will help,\nUltimeaciax
5946
The instruction is not correct because the total volume yielded when you mix 100ml of concentrated solution and 900ml water will be somewhat less than 1000ml. The concentration will be slightly higher than 1.0M\n\nThe correct procedure is to take a 1l volumetric flask  and add the concentrated solution to about 800ml water, then fill up to the mark.
371289
According to Sigma-Aldrich the solubility of MgCl2.6H2O is 1M at 20C (or 20.3030 g/100ml solution), so it shouldn't be possible to prepare 4.5 M MgCl2 in just water. If there are more solutes involved which might increase the solubility of MgCl2 try dissolving those first and then the MgCl2.\n\nYet I saw some papers using 4.5 or 2.5 M...\n\nI don't know why there is this discrepancy between sigma-aldrich and the solubility provided in the link in the above answer which would correspond to 8 M if you consider 100ml of solution and not water. In reality you would have a significant change in volume when you add so much salt and the solubility would be much less but I don't know exactly how much. If you consider 167g of MgCl2.6H2O then you are adding 88.8 g of water which means that the total volume is at least 189 ml and thus the maximum concentration is 4.3 M.\n\nWikipedia gives a different number; 54.2g/100ml at 20C, without specifying if it is for the anhydrous or hexahydrate... so it would be 5.6 or 2.6M respectively... \n\nWhat a mess !!!\n\nMaybe you should also post the question in biology, since someone who has been doing such purifications can clarify the issue.\n\nAnother idea would be to prepare a 100ml saturated solution according to the lowest reported solubility (that of sigma-aldrich) using a 250 ml volumetric cylinder, and then from a pre-weighed amount of MgCl2.6H2O keep adding powder slowly until it doesn't dissolve any more. Weigh the powder that is left in order to find how much you dissolved, measure the final volume and calculate the molarity. If you achieve higher than 4.5 M, dilute.\n\nOR\n\nprepare a 4 M if the amount of powder that you couldn't dissolve is very small.\n\nor \n\nprepare 450ml of 1M and evaporate to 100ml of solution while praying that there will be no precipitation...\nThat's very desperate but I am running out of ideas...\n\nI must say the whole story is very fishy (finding very different solubility values)\n\nI'd go mad too...
473024
The product of a neutrilization reaction is always water and a salt, so use that as a guide.  For part 2, use the units as a guide.  Hint M=molarity=mols/L
510592
You would have 36.5 g of HCl in 500mL of the acid solution.\n\nThe easiest formula for dealing with a titration is:\n\nV(acid)*M(acid) = V(base)*M(base)\n\nWhere V is the volume in liters and M is the molar concantration (moles/Liter)
448049
Yes, it is true that the density of a solution is different of solvant but tjis difference is small ; So , i propose to dissove 40g of amyl acetate in a liter of water if that acetate is a solid. If that acetate is aliquid the total volume will be 1.04 l and you must use 47 cc tp water
139178
The amount is diluted by a factor of 25.  So the resulting solution would have a concentration of 11.6M divided by 25 or \nor 0.464M.
404360
All right, you know what M means, right?  It's short for moles per liter.  You want to have 0.250 moles of Al2(SO4)3 in each liter of solution.  You're preparing 0.4 L of solution, so you need (0.4 * 0.250) = 0.1 moles of Al(SO4)3.\n\nNow the molar mass of each element is listed on most periodic tables.  That's how much one mole of that atom weighs.  A molecule weighs as much as all the atoms in it put together, so a mole of Al2(SO4)3 weighs as much as two moles of Al, 3 moles of S and 12 moles of O.  That's 2(26.982) + 3(32.06) + 12(15.999), according to my periodic table.  So if you add it all up, a mole of Al2(SO4)3 weighs 342.132 grams.  You only need a tenth of a mole, so you use 34.2132 grams.
536609
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
414087
try asking this in the chemistry section next time....
419731
well I think no. because dilute HCl has a low concentration therefore it produces less hydrogen ions. so the less hydrogen ions it produces the less pH  it has
476263
you should learn how to do your own homework hun you'll never be a chemist with someone else doing your work.\n\n\nuse the molar mass from the periodic table (units = g/1mol)\n\nCl = 35.5\nH = 1.00\n\nMolar mass, formula weight, molecular weight, formula mass = 36.5 grams HCl/1 mol HCl\n\n\nThis is a simple problem is you can't do this problem maybe you should try changing majors.  I hear history or english is nice.
420051
yes u r correct , u need vapour pressure of H2O. From vapour pressure calculate vapour density of H2O\n  Now, u can easily calculate molecular mass of the unknown as it  is twice the vapour density
181824
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.
464673
1: You wouldn't measure a dry powder in milliliters.  You can't really know exactly how much water to add unless the know the mls the dissolved drug take up.  You could add 10mls then QS up to 15 mls once it dissolved in the orginal 10mls.\n\n2 If the total volume of 2 grams is 15mls.  Then 7.5 mls added to 50 mls of NS or D5W would be the answer.\n\n3.  If the total volume is 100 mls then 100ml/60 minutes= 1.66ml/min. Of course, you would set up the pump in mls/hr which in this case is 100ml/hr...duh.  If the total volume is 107.5 mls then 1.8mls/min.
247376
I don't know if this is right ... but check it and tell me\n\nV*C KOH = V*C HBr\n\nV KOH = 18.6ml C KOH = 0.752M\nV HBr = 22.1ml C HBr = ?\n\n18.6*0.752 = 22.1*C HBr\n\nC HBr = 0.6329M\nIts seems reasonable cause the difference between concentrations isn't significant
520035
molarity (M)=(weight/mol.weight)*(1000/volume of the solution in mL)\n\nNow M = (27/17)*(1000//755)=2.1036 mol/lit