Calculate the molarity of 20.00 L of solution that contains 100.0 g of NaOH.?
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
313620
M(conc.base) x vol.(conc.base) = M(dil. base) x vol.(dil. base)\n12 x vol. = 2.5 x 350\n vol = 72.9 ml \n\nIn a graduated cylinder, take 72.9 ml of concentrated NaOH. And in a standard flask, take 350 - 72.9 = 277.1 ml of water. To this flask, add the concentrated NaOH (72.9ml) from the graduated cylinder and shake the flask for uniform concentration. Now you've prepared 350 ml of diluted NaOH solution.
901225
I don't do homework but I'll tell you how to do it.  I suspect, though, that you mean 0.9% (w/v).  The parenthetical expression means that the 0.9% is in terms of weight to volume.\n\nSo, for every 100 mL of this solution, you have 0.9 g NaCl.  That means that, for every liter, you have 9 g NaCl.  So, you calculate how many moles of NaCl are in 9 g of it.  Then, since you know that's in one liter, that will be your molarity as well.
773833
2NaOH + H2SO4  = Na2SO4 +2H2O                                      Sodium Hydroxide is 0.5moles.Two Moles neutralise One mole of Acid.   0.5Moles neutralise of Alkali neutralises 0.25Moles of Acid                                                                                           50mls of 1M SULPHURIC ACID = .05M THEREFORE MOLAR CONCENTRATION IS 5
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
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).
715055
First off, you need the concentration of the NaOH solution.  From that you can determine the number of moles of NaOH that were used.  This is equal to volume (V) time the concentration, in mol/L (C).\nn = V * C = 0.02260 L * ? mol/L = ? mol\nn = 0.02200 L * ? mol/L = ? mol\n\nThe reaction of KHP (C8H5O4K) and NaOH is this:\nHOOC-C6H4-COOK + NaOH -> NaOOC-C6H4-COOK + H2O\nSo, 1 mol of NaOH will react with 1 mol of KHP.\nThat means that the number of moles of NaOH you calculated above is also equal to the number of moles of KHP.\nThe molecular weight of KHP is 12*8 + 1*5 + 16*4 + 39 = 204 g/mol\nMultiply that by the number of moles: 204 g/mol * ? mol = g actual KHP\n\nYou can divide that number by the 0.911 and 0.950 g and then multiply by 100% to find the purity.\n\nHere’s an example if your NaOH solution was 0.1 M (M means ‘molar,’ or mol/L):\nn = 0.02260 L * 0.1M = 0.002260 mol of NaOH = 0.002260 mol of KHP\n204 g/mol * 0.002260 mol of KHP = 0.461 g pure KHP\n0.461 g pure KHP / 0.911 g impure KHP * 100% = 50.6 % KHP\n\nHope this helps!
813561
The molecular mass of NaOH is 40.00 gm/mol,  so 8 gm of NaOH contains 8 gm/40.00 gm.mol = 0.20 moles of NaOH.\n\nThe molecular mass of HCl is 36.46 gm/mol, so 8 gm of HCl contains 8gm/36.46 gm/mol = 0.22 moles of HCl.\n\nIf we react these amounts of NaOH and HCl, the NaOH will be the limiting reactant, and we will produce 0.20 moles each of NaCl and H2O.  This reaction will liberate \n\n0.20 mol * 56 kJ/mol = 11.2 kJ of energy as heat\n\nThe heat capacity of the final solution is given by the mass of the final solution (which, by conservation of mass, is equal to the mass of the initial solutions) multiplied by the specific heat capacity of the final solution.  With the assumptions given in the question, the heat capacity of the final solution is:\n\n(58 gm + 258 gm) * 4.184 J/(gm*degC) = 1322 J/degC.\n\n(Note that we don't need to use the density of the final solution.)\n\nThe temperature change of the solution is given by the energy released divided by the heat capacity of the system:\n\n(11.2*10^3 J)/(1322 J/degC) = 8.47 deg C\n\nIf the initial solutions had a temperature of 25 C, the final solution has a temperature of 25 + 8.47 C = 33.47 C.
927975
15%(m/v) means 15grams of NaOH in 100ml of solution.\nHence 120*100/15= 800ml will contain 120 g of NaOH
964051
First you must know the equation of the reaction, so you can establish the mole connections between the involved substances\n\n2HNO3(aq) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2H2O(l)\n\nNow, you have 60,0 mL of 0,100 M Mg(OH)2(aq)\nTherefore:\n0,100M*(60,0 mL)(1L/1000 mL)= 6,00*10^-3 moles of pure Mg(OH)2\nNow, 2 moles of HNO3(aq) react with 1 mol of Mg(OH)2(aq)\n\n6,00*10^-3 moles Mg(OH)2*(2 moles HNO3/1mol Mg(OH)2 )= 0,0120 moles of HNO3(aq)\nMolarity = moles solute/volume of solution\n0,200M= 0,0120 moles /?\n?= 2,40*10^-3 L of solution= \n2,40 mL of 0.200M HNO3(aq)  are needed to completely neutralize 60.0 ml of 0.100 M Mg(OH)2 solution
700313
NaOH + K+ = Na+ + KOH
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.
320061
Someone else has asked this same question before, and it really isn't that hard . . .\n\n60 ml of 0.250M HCl plus 60 ml of 0.125M NaOH will result in 120 ml of solution total.  All of the NaOH will have been neutralized by the HCl, leaving an excess of HCl, so the solution will be acidic -- the pH will be much less than 7.\n\n60 ml * (1.0 liter/1000 ml) * (0.250 mole/liter HCl) = .015 mole HCl\n\n60 ml * (1.0 liter/1000 ml) * (0.125 mole/liter NaOH) = .0075 moles NaOH\n\nThat means you will have 0.015 - 0.075 = 0.075 moles of HCl left over after the neutralization.\n\nI leave it for you to calculate the pH of a solution of 0.075 moles of HCl in 120 ml of solution.  Remember that pH = -log(base 10)[H+]
802376
YOUR ANSWER IS CORRECT!  Here's how i did it:\n\nYou can find the # mol of NaOH used:\n\nMolarity = (# mol)/(volume solution)\n0.4 = (# mol)/(.075 L)\n\nSolving, you get 0.03 mol of NaOH.\n\nUsing the balanced equation:\n\n3 NaOH + H3PO4 ==> Na3PO4 + 3 H2O \n\nYou see that 3 mol NaOH = 1 mol H3PO4.  Thus, if we have 0.03 mol NaOH, that means we have 0.01 mol H3PO4.\n\nNow we can solve for the molarity:\n\nMolarity = (0.01 mol H3PO4)/(0.05 L) = 0.2 M
906606
1) To prepare this solution you should do:\n\n100 g of solution needs 2.5 g of solute (to get 2.5% mass/mass)\n150 g of solution needs  x    g of solute (to get 2.5% mass/mass)\n\n2)  x = (150 g x 2.5 g / 100 g) = 3.75 grames of KCl\n\nThat's it!\n\nGood luck!
685497
Molarity = moles/volume, so volume = moles/molarity\n\n18.5 grams NaOH /40.0 g/mol = 0.4625 moles\n\nVolume = 0.4625/0.750 = 0.6167 L = 616.7 ml
1051997
This is the most handy of chemistry equations, if you need to take more chem, commit it to memory:\n\nm1v1=m2v2, molarity*volume=diluted molarity*diluted volume\n\nSo, 9.9M*.102=x*.525.  Then solve for x by dividing the left side by .525.  You will get the concentration in mol/L of the final solution.\nSorry, I read your question wrong, I read it as diluted to, not diluted with, so you just need to add the volumes.\nSo it's 9.9*.102=x*(.102+.525)\n9.9*.102/.627=concentration of 1.6\nSorry, I wasn't more carefull.\nThat really is the way to solve these equations though.