What is the molarity of this solution ?
914609
9,50 g HCl*(1mol HCl/36,45g HCl)= 0,261 moles HCl\nM= 0,261 moles/0,0292L\n= 8,93 M
859821
2.2 gram\nu can calculate it by the molarity formulae
837288
Use the formula M1V1=M2V2\n\nThat is the first Molarity times the first volume are equal to the second Molarity times the second Voume.\n\n.2098M x70mL = .662M x V2\n\nSolve for V2.\n\n(.2098 X 70)/.622 = V2\n\nV2 = 23.61mL NaOH to titrate the solution to neutrality.
445032
Click on "Add details," post the problem, and, as SMeghead said, someone might be able to help.\n\n(EDIT: You didn't provide too much information. From what you gave, about all we can say is that 12.5% of the solute - 1% from 8% - evaporated. If you need to know how many grams, milliliters, cubic centimeters, etc. evaporated, we'll need more information.)
245982
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
1104578
Use the equation M1V1 = M2V2\n\nM1 is the molarity of the HCL or 9.9 M\n\nV1 is the first volume of 102 ml but it needs to be changed to Liters. So it is 0.102L\n\nV2 is the total volume of the final solution so 525 ml (0.525 L) + V1 or 0.102L = 0.627\n\nM2 is what you are looking for - the final concentration or molarity.\n\nRearrange the equation to ((M1)(V1)) / (V2)  = M2\n\n(0.102L)(9.9M) / (0.627) = 1.61 M\n\nHope this helped!
482555
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit
298095
If the freezing point of the solution had been incorrectly read .3̊ lower than the true freezing point, then the observed ΔTf would have been greater than it should have been. This would have caused the calculated molality to be too high. If the calculated molality were too high, then the number of moles present as calculated would be too high. An\nexaggerated number of moles would cause the calculated molar mass to be too low, as the same weight of the substance would be distributed against more moles of that substance.\nThe substance would seem lighter than it is.\n\nI'm sure this will help,\nUltimeaciax
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.
807846
Molarity is mol per litre of solution.\n\nAssuming the volume of salt is negligible, we just add 1 mol of NaCl to 1 litre of water.\n\nTo find 1 mol of NaCl, add the molar mass of each substance.\n22.99 +35.45 = 58.44g\n\nSo, you add 58.44g of salt to 1 litre of water.\n\n(If you want to take into account the volume of salt, find the density to determine the volume, and subtract that volume from 1L of water to determine the amount of water to add, though generally we just assume the volume is negligible)\n\nNOTE: This is only a 1 molar solution for the temperature and presure at which you made it. The problem with molarity is that it is mol per litre of solvent, and the volume of a substance can vary with changes in temperature and pressure.
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)
905387
Molarity is simple.  The formula is Moles of Solute divided by Liters of solution. \n\nThe solute is the smaller amount, if you get confused with the solute and solvent.  Many times the solvent is water.
601420
Well, I'm not doing your homework for you, but Molarity is Moles per liter.  So that means you would have to change grams to moles, hopefully you know how to do that.
848062
what does molarity mean LMAO lol
646163
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
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.
261798
The aqueous solubility of AgNO3 at 20C is 67.8 (solubility is 100*m2/(m1+m2) where m2 = mass of solute and m1 = mass of solvent).\n\nRearraging the equation gives us m2 = (m1*67.8/100)/(1-67.8/100).  or\nm2 = (m1*.678)/(1-.678) = 210.56g of AgNO3 @ 20C\n\nAt 50C the aqueous solubility of AgNO3 is 79.2 so\nm2 = (m1*.792)/(1-.792) = 380.77g of AgNO3 @ 50C\n\nTo find the amount needed to make the second solution saturated we must subtract answer 1 from answer 2.\n\nWe find that 170.21g of AgNO3 must be added to make the solution saturated at 50C.