please help, how many milliliters of a .100 solution of NaOH would be needed to contain 2 g of NaOH?
820326
If this is in Normality;\n\n0.1 Normal solution of  NaOH\n2 grams of NaOH  in a liter = 0.05 N\n\nN = weight in grams divided by MW (molecular weight ) = 40 / valence =1 / 1 liter. = 0.05\n\nMW = (Sodium) 23 + 16 (Oxygen) + 1 (hydrogen) = 40\n\nTherefore: Using the formula N1V1 = N2V2\n(0.1)( X) = (0.05) ( 1 liter)\n\nBy cross multiplication:\n\nX = (.0.05) ( 1) divided by ( 0.1) \nX =  0.5 liters or 500 milliliters\n\nbecause  1,000 ml = 1 liter\n\nYou will get the same answer for Molarity because the valence is one.\n\nso the answer is 500 milliliters.
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
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...
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
621533
NaOH ----------------> Na+   +   OH-\nFind the number of moles of NaOH\nMoles of NaOH = 0.10M * 0.150L = 0.015 moles\nSince the coefficients in fornt of NaOH, Na+ and OH- are all 1, therefore if 0.015 moles of NaOH disssociates, 0.015 moles of Na+ and 0.015 moles of OH- are produced.\n\nSimilarly we will work with HCl\nHCl --------------------> H+      +      Cl-\nMoles of HCL = 0.30M * 0.05L = 0.015 moles\nThis means the 0.015 moles of H+ will be produced\n\nSince H+ and OH- have equal amiunts, they will completely neutralize each other leaving the solution at neutral pH. Since the neutral pH is 7, therefore, the pH of this solution will be 7.
803943
Number of moles of acid = volume (in litres) x molarity =0.0182 x 0.1  =  0.00182 Moles\nThe equation for the reaction shows twice as many moles of acid is needed as carbonate\n\n2HCl  +  Na2CO3  =  2NaCl +  H2O  + CO2\n\nSo the number of moles of Na2CO3 =  0.00182 divided by 2 = 0.00091\n\nThis is how many moles are in 10cm3 of solution so in a litre there would be 100 x as much  =  0.091 so the concentration is 0.091M\nConcentration is also measured in g/litre and in this case it is 106 (the RMM of sodium carbonate) x 0.091 = 9.646g/litre
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
657714
NaOH + HCl --> NaCl + H20\n\nn(NaOH) = 1.00 x 32.6cm3\n        = 1.00 x 32.6 x 10^-6\n        = 3.26 x 10^-5  mol\n\nn(HCl) = 1.00 x 25cm3\n       = 1.00 x 25 x 10^-6\n       = 2.5 x 10^-5  mol\n\ntherefore, NaOH is in excess.. hence the final solution is alkaline, which means the pH is more than 7.\n\nDue to the limitation of my knowledge, i can't give you the exact pH value. Just hope that it will help you to continue solving...
364108
Hope you didn't turn in your homework based on the friendly_doc answer!  It's wrong.  1% means 1g per 100 mL.  Therefore, in 200 mL, there will be 2g.  So the correct answer to the question is 2.5 grams, or 2,500 mg.
780228
HCL and NaOH are 'strong' acid and base. When they react, the product is neutral. NaOH + HCL --> NaCl + H2O\n\nWhat you need to find is which is the limiting reactant, the other one (what is left over) will effect the pH by its concentration in the otherwise neutral solution.\n\nFor your first problem:\n\n0.0557M HCL = 0.0557mols/liter*0.00542L = 3.02*10^(-4)mol HCL\n\n0.0377M NaOH = 0.0377mols/liter*0.00337L = 1.27*10^(-4)mol NaOH\n\n3.02*10^(-4)mol HCL - 1.27*10^(-4)mol NaOH = 1.75*10^(-4)mols HCL\n\n0.00542L + 0.00337L = .00879L\n\n[(1.75*10^(-4)mols HCL) * (1 L) ] / (.00879L) = .020M HCL\n\npH = -log(.02) = 1.70
913013
0.9% NaCl means 0.9 g NaCl in 100 mL SOLUTION, so you need 9 g NaCl for 1000 mL (1L)\n\nTake a volumetric cylinder, putsome water in it, dissolve the salt and add water to a final volume of 1L and mix. If you just add 9 g of NaCl in 1L of water the final volume of the solution will be different than 1L and thus your concentration will deviate from 0.9%.
202477
I'm not sure if I understand your question.\n\nIf you are wanting to make a 0.92 M NaCl solution out of 0.39 M NaCl, well you are going to have to work hard.  Basically, you would be boiling the water and having to measure the mass of the flask.  This is a very tedious process.  Bottom line, you cannot use the dilution formula to make it happen.\n\nNow, if you are wanting to make a 0.39 M NaCl solution from a 0.92 M NaCl solution, then you use:\n\nM1V1 = M2V2\n\n0.92 * V1 = 0.39 * (.1)\n\nremember .1 L is 100 ml.\n\nSolve for V1\n\nV1 = 0.42 L\nV1 = 42 mL\n\nSo, take 42 ml of the 0.92 M NaCl and dilute it up to 100 ml.\n\nHope it helps!
929350
Do you mean M instead of N?\nIf you mean M, then\n1,00M/L*0,500 L= 0,500M\n0,500 moles NaOH*(40g NaOH/1 mol NaOH)= 20,0g of NaOH\nIt is c).
927975
15%(m/v) means 15grams of NaOH in 100ml of solution.\nHence 120*100/15= 800ml will contain 120 g of NaOH
907422
Since Na+ and OH- ratio is 1 to 1 in the solution, the concentration of OH- would be 0.1M as well. \n\nC = n x V\n\nNumber of Na+ ions is equal to the number of OH- ions. So n is both same. If the volume is fixed, the concentration should be 0.1M.
527028
This time i ans ur question but next time plz do ur homewrk ur self.\n\n1.\nWeight of NaOH = (4*.05) = .2 grams\nMoles of HCl =  Moles of NaOH = (.2/40) = .005\nVol. of HCl = .005*22.4 = .112 l = 112 ml\n\n2.\nWell let the grams be x\n1000x = 74.5*2.25*(75+x)\n(1000-167.25)x = 167.25*75\nx = 14.98 grams
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