plz help

[butterfly]

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

[Guest honeybabes_18]

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...

[-Demosthenes-]

wait, is is molarity of molality?

[Skye]

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.

[butterfly]

thankyou everyone it was a great help

[Guest brekka]

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?

[butterfly]

you multiply the concentration by the whole mass. then you devide the mass of 3CL by the whole mass, to get only anions.

[Guest sureksha]

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...............

[rogue]

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...............

 

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