msu08

10+ Year Member
7+ Year Member
Nov 14, 2006
60
0
Status
Pre-Dental
anyone want to try and explain these??

1) Normality of a Kpermanganate solution is 2.10. what is the molarity of the soln if in the course of the reaction the potassium permanganate is changed to manganese dioxide? answer is .7

i understand normality, and see that 2.10 is divided by 3, to get .7 but why 3?

2) If 3L of .5M NaCl is mixed with 9.0 L of .2777 M of NaCl what is the final concentration of the resulting soln? answer is .33M

3) How many grams of Al2(S04)3 are needed to make 87.5g of .3m Al2(SO4)3 soln? atomic weights of Al= 27, S=32, O=16
answer is (87.5 x 102.6)/1102.6

4) What is the mole fraction of C2H5OH in an aquesous soln that is simultaneously 3.86m C2H5OH and 2.14m CH3OH?
answer is 3.86/61.55

any help is GREATLY appreciated!! thanks!
 

Panther85

10+ Year Member
Aug 20, 2007
99
1
Status
Podiatry Student
1. yea its 3... NaCl breaks into Na+ and Cl-... so its 2. K+ MnO4- was at "2" but the problem states that it was now MnO2 which is 3 particles im guessing

2. such a tedious problem. ok u cant just add the M concentrations. u need to find the total moles of EACH solution separate (1.5 and 2.5). then add them together! thus u get 4... take 4 and divide it by the 2 total volumes (9+3=12) 4/12=.3333
 

Danny289

Member
10+ Year Member
Dec 2, 2006
1,523
4
Status
Pre-Dental
1- you know the normality formula N=M X n
n is different in bases and acids ( the number of "OH" or "H") in oxidation reduction is the number of electron change:
KMno4=======> Mno2
for Mn
Mn+7+ 3e ====> Mn+4 ====> that is "3" you are looking for.

2- is already answered.

3- for this question find out how much Al2(SO4)3 you need in 87.5g, 0.3 sol.
in other hand change the problem to this.
"how many grams Al2(SO4)3 in 87.5 g of 0.3M solution?"
(87.5/1000)X 0.3 X 342
4- in mole fraction remember water (1000gr water)
1000/18= 55.5
now add all the moles: 55.55 +3.86 + 2.14= 61.55
3.86/61.55 for C2H5OH
 
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