Watabe wa kemia msaada tafadhali..! | JamiiForums | The Home of Great Thinkers

Discussion in 'Jukwaa la Elimu (Education Forum)' started by b5-click, Jun 18, 2012.

1. ### b5-clickJF-Expert Member

#1
Jun 18, 2012
Joined: Mar 21, 2012
Messages: 1,969
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Wandugu p0leni na majukumu...
Kwa yey0te anaye weza kunipa msaada ktk swali hili afanye hivy0 tafadhali...
''If an aqueous solution boil at 100.5'C, at what temperature does it freeze? (For water Kb= 0.512'C/m and Kf= 1.86'C/m)...
Shukrani kwanza.!

2. ### Good GuyJF-Expert Member

#2
Jun 18, 2012
Joined: Oct 24, 2010
Messages: 4,511
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dah long tyme but ckupata gamba la kemia advance bhure,
ok for the boiling process
dT=Kbxconstants
but dt=(100.5-100)=0.5(since the liquid z water b.p=100)
hence
0.5=0.512xC
hence making C subject
c=0.5/0.512
after dat twende kwenye 2nd step(the freezing process)
dT=Kfxconstant
dT=KfxC
dT=1.86x(0.5/0.512)
but
dT=0-freezing point
so (again since ts water m.p=0)
then
0-[1.86x(0.5/0.512)]=freezing point(u should gt -ve somethng)
sina calculator so utajaza. Hiyo constant ni bunch of masses nmeshazisahau,ila zote zinabaki constant throughout(ts an assumption ofcourse)

3. ### VitendoJF-Expert Member

#3
Jun 19, 2012
Joined: Oct 23, 2009
Messages: 597