Calculate the molarity of a solution....?
928965
molarity  = moles / liters\n\nmoles = given mass / molecular mass\n\n10 g H2SO4 / 98 g per mole = about  0.1 moles\n\n0.1 moles / 1 L = 0.1 M solution
741771
0.5M = 500mM\nv x c=v'xc'\nv x 500 = 50 x 1\n0.1mL of the 0.5M solution diluted to 50mL will be 1mM
379843
Using the formula for pH: pH = -log(H+)\nwe can find the molarity\n\nFirst, we plug in what we know\nThe pH is 2.65\nand the H+ is the concentration of hydrogen ions measured in moles/ liter. Hey, that's what we're looking for!\n\n2.65 = -log(moles/liter)\n\nusing algebra we can solve for moles/liter\n\n-2.65 = log(moles/liter)\n\nLet's raise 10 to the power of each side\n\n10^(-2.65) = 10 raised to the negative 2.65 power\n10^(log[moles/liter]) = 10 raised to the power of log(...)\n\nOk, so..\n\n10^(-2.65) =10^(log[moles/liter])\n\nNow 10 with log as a power cancel each other out so...\n\n10^(-2.65) = moles/liter\n\nmoles/liter is the molarity, let's call it "M"\n\nM = 10^(-2.65) \n\nM = approx. 3.16 x 10^(-3)
795448
normality = molarity x ionizable H\n\n18 M x 2 H = 36 N
165799
You want us to calculate acitivites?\n\nYou need to determine the ionic strength and atomic radius in order to do that.
268708
Wow, this question is harder than I thought lol, I thought it would just be a plug and chug but its not u actually have to think!  I havent quite figured it out but I will give you a start:)  Maybe I'll come back to it.  \n\nSo the formula is: \nOsmotic Pressure in atm. = Molarity ( R ) (Kelvin Temp.)\n\nOsmotic Pressure = 3.05 atm\n\nR = 0.0821 liter-atm/mole kelvin \n\nConvert the temperature from Celsius to Kelvin. \nK = 10 + 273 = 283 K \n\nWe dont have the Molarity so first find that, then since we know Molarity = (g/ molecular weight)/liter and in the problem we have the relationship 4g/liter and 2g/liter.  Figure it out!:)\n\nGood luck!
968320
here, you just use the ideal gas law, PV=nRT.  since molarity is equal to moles of solute per liters of solution, M=n/V, so P=MRT.\ntherefore,\nKp=Pco*Ph2o/(Ph2*Pco2)\n  =[CO]RT*[H2O]RT/ ([H2]RT*[CO2]RT)\nall the RT coefficients end up canceling out, so:\nKp=[CO][H2O]/[H2][CO2]\n  =Kc\nsince there were equal numbers of gas molecules on each side of the equation, Kp was equal to Kc.  as a general rule, Kp=Kc*(RT)^N, where N is the change in the number of gas molecules.
936884
Mstock*Vstock=Mdil*Vdil =>\n\nVstock=(Mdil/Mstock)*Vdil = (0.04/Mstock)*500 mL\n\nSo when they give you the molarity of the stock solution you just plug it in the equation and get the volume (in mL) of stock that you need to use.
773889
Suspected trick question.  The volumes included are irrelevant.  You know the molecular weight (66000 Da) and you know how many micromoles you need total in each solution.  What does it matter if the one micromole is in 1 ml or 10K gallons?  You know it only contains one micromole.  Divide the molecular weight by 10^6 and you're done.  (or you have not stated the problem correctly)
948311
1) Calculate the number of moles of HCl:\n3.75/(1.008+35.45) = 0.103 mol\n\n2) Calculate the molarity, that is the number of moles per liter:\n0.103 mol / 20.2 x 10^-3 l = 5.1 mol/liter
768648
Actually it is not a solubility problem, but a buffer.\n\nLet's use the Henderson-Hasselbalch equation for simplicity\n\npH=pKa+log[CH3COONa]/[CH3COOH]\n\nIf V is the volume of the solution in L then\n[CH3COOH]= mole/V= 0.010/V\n[CH3COONa]= mole/V =x/V\n\nso substitute in the equation and you get\n\n5.37=4.74+ log((x/V)/(0.01/V))=>\n0.63=log(x/0.01) =>\nx/0.01= 10^0.63 =>\nx =0.01*10^0.63= 0.0427 mole =42.7 mmole or since MW=82\nm= 0.0427*82= 3.50 g
827418
First, you have to calculate the mass of each component by multiplying their volumes by their densities.  You should now have weights for the ethanol and for the water.\nNext, calculate the number of moles of each solvent by dividing those weights by their atomic weights.\n\nNow you can begin answering the questions.  Molarity is defined as the number of moles of a substance per liter of solution.  The total volume of the solution, due to the assumption that the volumes are additive, is 100 ml + 200 ml, giving you 300 ml or 0.3 liters.  So, to get the molarity with respect to ethanol, divide the number of moles of ethanol by 0.3 L.\n\nMolality is the number of moles per kilogram of solvent.  Do exactly the same as above, except this time, you have to calculate the total Kg of solvent instead of the total volume of solvent.  You have to add the mass of ethanol and the mass of water you have calculated from the volumes and the densities.  That's your total mass but don't forget to divide by 1000 to get Kg.  Then you calculate the molality of ethanol by dividing the number of moles of ethanol by the weight of your total solution in Kg.\n\nThe mass percent of ethanol would be the mass of ethanol divided by the total mass of the solution multiplied by 100%.  This is just like any other percentage calculation.\n\nThe mole fraction of ethanol can be calculated by dividing the number of moles of ethanol by the sum of the moles of ethanol and water.\n\nGood luck.
1186593
Mix with a 0% solution in the relation 33:2\n\nSo use 2 units of your solution (like 2 liters) and 33 units of the 0% solution (like 33 liters).\n\nThat's all.
727216
use beers law to figure out the concentration and then take that to mole ratio
921065
you know you have got 152g total. So 1.8g/ml means you have got 152/1.8 for volume.Now just use MW KOH. Molarity will be moles KOH per litre, molality will be moles KOH/kg of solvent and Normality in this case will be the same as molarity but written  with symbol N.\n\nAs an aside the density of this solution would not be anywhere near 1.8g/ml, are you sure you have this right?
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.
633715
5. KHCrO4 ----------------> K+   +   HCrO4-\nSince KHCrO4 dissociates complteley,\n the molarity of HCrO4- = molarity of KHCrO4\n the molarity of HCrO4- = 0.05M\n\nNow HCrO4 ------------------>    H+    +     (CrO4)2-\nCalculate the concentration of H+ from the pH value\n[H+] = 10^(-pH)   = 10^(-3.80)  = 1.58 * 10^(-4)\nIt is clear the only source of H+ is HCrO4-, therefore the concentration of (CrO4)2- is the same as concentration of H+\n[(CrO4)2-]   =   [H+]   = 1.58 * 10^-4\n\nKa = {H+} * [(CrO4)2-] / [HCrO4-]\nKa = [1.58 * 10^(-4)]^2/0.05\nKa = 5.02 * 10^ (-7)\n\n6.       KOH----------------> K+    +    OH-\n         HClO4 -------------> H+    +    ClO4-\n(a) \n    Moles of KOH = Molarity * Volume = 0.200M * 0.030L\n    Moles of KOH = 0.006 moles\n    From the equation above, \n    Moles of OH- = moles of KOH = 0.006 moles\n    \n    Moles of HClO4 = Molarity * Volume = 0.150M * 0.030L\n    Moles of HClO4 = 0.0045 moles\n    From the equation above,\n    Moles of H+ = moles of HClO4  = 0.0045 moles\n\n     H+    +    OH-    ------------>   H2O\nWhen this reaction is finished, 0.0015 moles of OH- are left\n[OH-]  =  moles of remaining OH- / Total Volume\n[OH-]  =  0.0015 moles/ 0.06L (30ml + 30ml)\n[OH-]  =  0.025M\npOH = -log [OH-]\npOH = -log (0.025) = 1.60\npH = 14 - pOH  = 14 - 1.60 = 12.40 \n\nDo the same for parts b, c, d, and e\n\n7. I will assume that \n   H2S --------------------------> 2H+    +    S2-\n   Ksp of HgS = 2.0 * 10^(-53)\n   Ksp of FeS = 6.3 * 10^(-18)\n\nKsp of HgS = [Hg2+] * [S2-]\n[Hg2+] = 0.01 M\n[S2-] = x\n2.0 * 10^(-53) = x*0.01\nx = 2.0 * 10^(-51)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 2.0*10^(-51) = 4*10^(-51)\npH = -log (4*10^(-51)) = 50.39\n\nKsp of FeS = [Fe2+] * [S2-]\n[Fe2+] = 0.01M\n[S2-] = x\n6.3* 10^(-18) = x* 0.01\nx = 6.3 * 10^(-18)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 6.3 * 10^(-18) = 1.26*10^(-17)\npH = -log (1.26*10^(-17)) = 16.90\n\nthe pH range is from 16.90-50.39\nThis is the best explanation I can gibe. By the way, pH level beyond 0 and 14 do exist, but, I'm not 100% about this answer.\nI fixed it, I think it is correct now.