How many grams of water are needed to prepare 250g of a solution that is 5.00% by mass NaCl?
479787
In 100g of solution that is 5% by mass NaCl, there are 95 g H2O and 5g Nacl.\nYou want 250g of solution...set up a simple ratio:\n95g H2O/100g Soln = ?g H2O/250g Soln\n? = 237.5g H2O
1171925
PH of the aqueous solution of NaCl is 6 - 7 @ 1% by weight
793702
Firstly,\nNaCl dissociates into Na+(aq) + Cl-(aq)\n1faraday of charge is required to produce 23 g Na from molten NaCl .\ntherefore production of 100000 g of Na will require charge = \n(1/23)*1000000 Faradays = 43478.26 Faradays = 43478.26 * 96500 C      [ as 1F = 96500 c]\n=4.1956 * 10^9 C\nBut, Q=IT \nso, T=Q/I = (4.1956 * 10^9)/300000 = 13.9855 * 10^3 secs = 3.88486 hrs.
603847
if its 2% by mass, thats 2 grams of KCl for every 100 grams of solution its in. \n\n2g KCl= 2g / (74.55g/mol)KCl = 0.0268mol KCl\n\nBecause there is one mole of Cl in KCl, this is also the number of moles of Cl in 100 grams of solution.\n\n0.0268mol Cl * (35.45g/mol)Cl = 0.951gCl/100gsolution\n\nThis is 0.951%\n\nnote: 35.45g/mol is the molar mass of Cl and 74.55g/mol is the molar mass of KCl. These were obtained from the periodic table.
984571
1 litre (1000 ml) of 1 M solution of NaCL contains 23 + 35.5 gms =58.5 gms (ie 1 molecular wt of sodium chloride in gms dissolved in 1 lit of solution)\n\nHence 1 lit of 0.360 M solution  will contain 0.360 X 58.5 gms or 21.06 NaCl (Sodium Chloride) \n\nTo find out exact vole in ml which contain 21.14 find volume by this formula 2.14 divide by 21.06 multiply 1000 , (ie 2.14/21.06 * 1000 ) 101.6 ml\n\nPS\nbasic assumption is 0.360 M indicates .360 Molar solution. \n(ie: M stands for Molar)
39876
To answer this question you must look at your given data. 13% of 1.1 = .143 so you have .143g NaCl for every ml solution. Molarity is moles solute/liters solution. You just calculate how many grams NaCl are in a liter and convert to moles, do the division and your done.
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...
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
1136961
Are you preparing a buffer soln with NaHCO3 and NaCl ? Then take equimolar amounts of both and dissolve them in 100 ml of water and then dilute them to the desired strength.
158276
25 gms is 0.245 moles which would require 0.7 L of 0.35M soln.
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
831841
Let g = grams of salt added.\n\nThere are already 3 g of salt in there.  So there will be 3+g grams of salt when you are done, and there will be 25+g grams of total solution.\n\nWrite a proportion:  (3+g)/(25+g) = 20/100 and solve
342141
ok well, the equation is balanced so you need equal moles of Ag and Cl to make the products. They give you the concentration and volume of Ag, so you can find the amount of moles by multiplying 0.104mol/L * 0.0272L = 0.0028288 mols of Ag. so 0.0028288 mols of Cl were there originally. and you have the volume of water, so you divide moles by volume and you get the concentration. 0.0028288 mols/0.025L = 0.113152 M. \n\nanother way to do it is the easy way, use c1v1 = c2v2. you have c1 = 0.104M, v1 = 27.2mL, v2 = 25mL, so you can find c2. you dont have to worry about units in this case so its easier. \n\nhope this helped. let me know if u need more help. laterz
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.
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
439952
You can find in "Chemical Engineer's Handbook", from Perry-Chilton, graphics for density of water + NaCl solution vs. %w/w of NaCl. The NaCl concentration is 3.5%
829104
Well, it's 0.205 moles/liter and you're making 1/4 liter, so you need 0.205/4 or 0.05125 moles.  The molar mass of NaOH is equal to the sum of the molar masses of Na, O, and H, or 22.990 + 15.999 + 1.007 = 39.996 g/mole.  The mass on NaOH needed is equal to 0.05125mole*39.996 g/mole = 2.050 grams.