A 5 g vial requires adding 8.6 ml diluent to get a concentration of 250 mg/mL. The powder volume is ______mL.?
913893
5000 mg / 250 mg/mL = 20 mL\n\n20 mL - 8.6 mL = 11.4 mL
938621
Ok I don't have time to do all of these, but --\n\n3 is 34 minutes 8 seconds\n4 is 2.5 metres\n5 is 9.80 metres\n6 is 50 grams\n7 is 18 x 1000 = 18000 grams\n8 could be 5 cubic centimetres (5 cc)\n  or could be 5 millilitres (5 ml) (it depends whether it's a liquid or a gel)
1140664
You're missing a unit. Do you mean 2.5 kg solution?
784999
.15 M solution of NaCl means .15 moles of NaCl (solute) in 1000g of water (solvent).\ntherefore, to make 250 ml (or g) solution [ as density of H2O is 1]\nmoles of NaCl required = .15/4 = .0375 moles \n& moles = given wt * molecular wt \nso, given wt (or required wt in this case)= moles/ molecular wt. = .0375/58.1 = 645.4388 * 10^-6 g
239849
.7M x 20mL = .25M x (Volume of new IA solution)\nVol = (.7 x 20) / .25\nVol= 56mL
341279
125mL of 0.251 M HCl gives 0.125 * 0.251 mols of HCl\n\nThen 250 + 125 mL is 375 mL = 0.375 L\n\nSo 0.125 * 0.251 mols in 0.375 L gives 0.125 * 0.251/0.375 M\n\nThat is 0.084 M
710835
Suppose that we have r milliliters of the 8% solution.  This means the amount of salt in it is .08r.\nSuppose we have s milliliters of the 3% solution.  This means that the amount of salt in it is .03s. \nWhen we mix them together we have r + s milliliters total, and the amount of salt in it is .08r + .03s.  If the resulting mixture is a 5% solution then this means that  \n(.08r + .03s)/(r + s) = .05.\nSo .08r + .03s = .05(r + s) = .05r + .05s.  \nThen .03r = .02s, or r/s = .02/.03 = 2/3. \nBut we know that s = 600 mL, so r = (2/3)s = (2/3)600 = 400 mL\n\nAdd 400 mL of the 8% solution to 600 mL of the 3% solution and get 1000 mL of 5% solution.
500934
Combine 19 parts of solvent with 1 part solute.  Boom, 5% solution.  \n\nIf you have an already diluted substance, just figure out the ratio and try to make it 1/20.  For example, if you have a 50% solution of NaCl and water, add two parts (2*50% is 100% or 1 part) of it to 18 parts water to make a 5% NaCl and H20 solution.
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
703531
FW of NaCl is 58.44  then\n\n   58.44 gm in 1000 mL = 1 Molar\n\n   4.78  gm in 200 mL = ?\n\ndo cross multiplication\n\n(4.78 gm x 1000 mL x 1 M) / (58.44 gm x 200 mL) = 0.4089 M\n\n(gm/gm and mL/mL cacels out so you get answer in M)\n\n\nNow you are taking 28 mL of 0.4089 M solution and diluting it to make final volume of 100.  Use M1V1 = M2V2\n\n0.4089 M x 28 mL = ? x 100 mL\n\n\nThen  ? = (0.4089 M x 28 mL) / 100 mL    { mL/mL cancels out and you get answer in Molarity)\n\n\n? = 0.1145 Molar is your answer
768198
molarity =moles/V where volume is expressed in L.\nmoles=mass/MW so\n\nmolarity=mass/(MW*V) =50/(106*0.5)=  0.94 M\n\nFor solutions, % (w/v) is grams of solute per 100 mL of solution.\nThus you have (50/500)*100% = 10%\n\n% is parts per hundred.\nso it is 10^4 times bigger than ppm\n so it is 10*10^4 ppm= 10^5 ppm.\n1 ppm corresponds to 1 mg of solute per 1L of solution.
679263
I think that what is troubling you could be an exerecise involving "molarity" which uses two quantities: mass(moles)/volume(liter). \nBut you have to bear in mind that mass is referred to the solute, and volume refers to the solution. So, these quantities are not referring to ONE substance. \nIt would be better if you posted a question with the whole exercise, for clarity. \nGood luck.
732197
The density can be assumed to be 1.0 g/ml.  \n\nSo you have 25 ml x 1.0 g/ml = 25 g H2O\n\nThe molar mass of water is 18 g/mol, so you have:\n\n25 g H2O/18 g H2O/mol H2O = 1.4 mol H2O
683322
Well, I suppose 1.25 g is 1250 mg, so seeing there's 1250 mg in each ml of solution, taking out 0.2 ml of the solution will net you 250 mg of the substance.
739325
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...
186466
A quantity of 22% acid solution is being diluted with some 8% acid solution. The final solution must be 12% acid & contain 101 mL. How much of each solution must be added to accomplish this?- write and solve a system of equations that describes this.\n\nx = amount of the 22% soln\ny = amount of the 8% soln\nx + y = amount of the 12% soln = 101 ml\n\n.22x + .08y = (x + y)*(.12)\n.22x + .08y = .12x + .12y\n.10x - .04y = 0\n\nx + y = 101\nx = 101 - y\n\n.10(101 - y) - .04y = 0\n10.1 - .10y - .04y = 0\n10.1 - .14y = 0\n10.1 = .14y\n(10.1)/(.14) = y\n72.143 ml = y\n\nx + y = 101 ml\nx = 101 - y\nx = 101 - 72.143\nx = 28.857 ml
742488
C1V1=C2V2\n40%x100mL=25%xV2\n4000=25%xV2\n4000/25=V2   ((25%xV2)/25=V2)\n160=V2\nV2-V1=60\nshe should add 60 mL of water.