Determine the Molarity (M, moles solute / Liter of solution) of a solution formed when 2.134E-3 kg Aluminum chloride is dissolved in water such that the final volume of the solution is 0.4356 kL.
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I know how to find Molarity but I'm not sure where the final volume comes in in the equation. Help? Thanks. :)
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Molarity is equal to moles of solute per liter of solution. The problem is giving you a mass of Aluminum chloride in kg, this is the solute of your solution. You have to calculate how many moles of Aluminum chloride are in 2.134E-3 kg Aluminum chloride. You will need the molecular mas of Aluminum chloride. Then divide this value by the volume of your solution (0.4356 kL) converted in liters. That will be the molarity of the solution
did you got the answer? what value did you get?
Ok I did (2134g AlCl3*1 mol AlCl3)/133.3405gAlCl3 = 16.004molAlCl3. I'm confused to where I go from here.
Would it be 16.004molAlCl3/435.6L = 0.03674 mol/L
Is my answer right? @dan815 @TheSmartOne @Nnesha @Luigi0210 @whpalmer4 @compassionate @abb0t @mathmate @KyanTheDoodle
the procedure is correct the only problem that I see is the conversion from kg to g 2.134E-3 kg = 2.134 g
I'm not that good in Chemistry, unfortunately ;-; @JoannaBlackwelder @chmvijay @Abhisar @matt101 @Australopithecus @alphadxg can assist you better in chem :)
@Cuanchi (2.134 g AlCl3 * 1 mol AlCl3) / 133.3405 g AlCl3 = 0.016004 mol AlCl3 0.016004 mol AlCl3/ 435.6 L AlCl3 = 0.00003674 mol/L
I got the same answer
3.67 x 10 ^(-5)
@Lena772 Just a guess, can you check if the volume of solution is 0.4536kL or 0.4356kL?