butterfly Posted April 9, 2004 Share Posted April 9, 2004 i am trying to work out these questions but i cant seem to get the right answer. any help we be appreciated. question one: Calculate the chemical amount of anions in 79.0 mL of a 0.4564 M solution of cobalt(III) chloride question 2: Calculate the concentration of anions if 100.0 mL of a 0.3530 M solution of iron(III) sulfate is diluted with water to the mark in a 500.0 mL volumetric flask Link to comment Share on other sites More sharing options...
Guest honeybabes_18 Posted April 9, 2004 Share Posted April 9, 2004 question 1. First step, calculate mass by multiplying concentration by molar mass... Then, follow the steps i gave you in the last post until you get to the last one where you divide by the amount... except this time you MULTIPLY by the amount (0.079L) That should give you the right answer Question 2. I have no clue... Link to comment Share on other sites More sharing options...
-Demosthenes- Posted April 9, 2004 Share Posted April 9, 2004 wait, is is molarity of molality? Link to comment Share on other sites More sharing options...
Skye Posted April 9, 2004 Share Posted April 9, 2004 Q2. This should work... Find the mol. wgt. of iron and sulphate, then find what percentage of their combined mol. wgt. is the sulphate, as it's the anion in this case. Multiply this by the molarity of the solution, 0.3530, to get the molarity of the sulphate. As this is diluted five times in the 500 ml flask, divide this number by five to get the conc. of the of the anions in the final solution, in mol/l. Link to comment Share on other sites More sharing options...
butterfly Posted April 10, 2004 Author Share Posted April 10, 2004 thankyou everyone it was a great help Link to comment Share on other sites More sharing options...
Guest brekka Posted April 10, 2004 Share Posted April 10, 2004 question 1. First step' date=' calculate mass by multiplying concentration by molar mass... [/quote'] See here, do you multiply by the molar mass of cobalt(III) chloride, or just 3Cl as it is only asking for anions? Link to comment Share on other sites More sharing options...
butterfly Posted April 11, 2004 Author Share Posted April 11, 2004 you multiply the concentration by the whole mass. then you devide the mass of 3CL by the whole mass, to get only anions. Link to comment Share on other sites More sharing options...
Guest sureksha Posted April 11, 2004 Share Posted April 11, 2004 can someone help me plzzzzzzzz.... i tried but i cant seem to get the rite ans for ques no 2 plzzzz save me.. got an assign due tonight............... Link to comment Share on other sites More sharing options...
rogue Posted April 11, 2004 Share Posted April 11, 2004 can someone help me plzzzzzzzz.... i tried but i cant seem to get the rite ans for ques no 2plzzzz save me.. got an assign due tonight............... Q2) Cint=0.3530M Vint=0.1 Cfin=? Vfin=0.5 n=Cint×Vint =0.0353moles so 0.0353moles=Cfin×Vfin Cfinal=0.0353moles÷0.5 Cfin=0.0706mol × 3 Final con.=0.2118mol/L Link to comment Share on other sites More sharing options...
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