Estimation of Halogens Video Lecture - NEET

hello friends this video on organic
chemistry b6 550 is brought to you by
exam clear calm no more fear from exam
oh let's see
estimation of halogens so we have
chlorine you have bromine you know ideal
I want to estimate the halogen so in
this case we use various method so in
this case a known mass of organic
compound is he did it might leak acid in
the presence of silver nitrate please
know a known mass of organic compound
this is my or getting along is it it is
heated with nitric acid in the presence
of silver nitrate so this is my H no.3
nitric acid this is my silver nitrate
and this is we're all getting compound
correctly so now in this case whatever
carbon hydrogen I had is there has PI
bond hydrogen in Tori so whatever carbon
hydrogen we have got it is oxidized to
carbon dies in water and we don't care
about it let it go right but we have
chlorine so we have AG no.3 this reacts
with my chlorine so I have AG no.3 this
reacts with any of the halogen so in use
egx
agx I have got right AJ's I have got
silver like you can easily filter it you
can dry it and then you can wait so I
have a G X you just filter it wash it
dried and wait so with that I can find
out percentage on hello so if you see
here the percentage of halogens will be
one atomic weight of halogen because
this is very always right by molecular
weight of my silver whatever you get a
GX into mass of a G X which is form
right by
mass of ordering come in 200 Y C you
should know the atomic weight of halogen
you should know the molecular mass
weight of this guy form because this guy
actually will give you the mass of
halogen the mass of halogen actually
correct and this is the mass of pollen
upon the mass of hydrogen total mass of
halogen found in the experiment by mass
of organic Evadne give you the
percentage of how much let me show you
actually so let's suppose M mass of
Aegeus was formal so I know that one
mole of one mole of X implies the live
one mole of D X so what is the molecular
weight of X so we'll write here
molecular weight of X implies more
liquid or weight of n GX correct so let
suppose the molecular weight of X is it
suppose 80 correct so 80 grams of X let
me as you will see bromine will give you
a G X will be now 188 money duty grams
of ATX correct so one gram of a G C will
be 1 so 1 gram of 18 X will be 1 1 gram
of Ag X if you get this implies
molecular weight of X by molecular
weight of a G's this much gram of a
correct and if you get em gram of em
gram of Ag eggs this imply this whole
thing into
correct so this is what it is nothing
but M gram of Henderson so percentage of
everything about the gram of halogen
divided by the weight of boiling upon
into 100 so you get this formula
correct
let me take with this value than one
it'll be clear so let me assume this is
my bromine so one mole of bromine is
growing molecular weight of bromine is
what 80 and molecular weight of a JX is
what one in here that means 80 grams of
bromine will give 180 degrees of silver
bromide
so that means 180 gram of silver halide
silver bromide agreed
this implies that we had 80 grams of
bromine in the organic upon so if you
got one grams of silver bromide that
means you had how much if it is 1 then
this is 80 by 180 correct so if you have
got one gram of Ag X that means you have
80 by 180 grams of silver in the sorry
brownie in the compound and if you have
you got em gram of Ag X that means you
have 85 one it it into M the RAM of
bromine in the oven come on all right
and this is what we had done here
let us see some examples say is that in
the Kiryas method of halogen 0.15 gram
of organic compound gave point 1/2 gram
of eg via that means I have my organic
compound so it gave me when I reacted
with actually
AG no.3 it gives me a GB are correct so
it has some bromine Ashley correct so
either you can directly use the formula
or you can just you
this fashion also so bromine the
molecular mass has what 80 and this guy
is easy one idea trying polygamous and
knocking balls
correct so you can say that 188 gram of
a GB Arak we have got this implies 80
gram of bromine in opening upon I can
write so if you've got 180 degree of
silver bromide this implies that there
was a degree of Romanian burning upon
but you have got how much you have got
point 1/2 gram of Asia via so let me
make it first one so if you have word
let's suppose one gram of a gbbr this
implies you have 81 188 gram of beer in
my opening upon correct but I have got
one month my gram if I got points only
fi got point 1/2 gram of eg BIA so point
1/2 gram of Ag beer if I've got in the
output this implies 80 by 188 into 0.12
gram of going so this is my gram of
growing in my organism bond you want to
find percentage percentage of domine
this is nothing but this whole thing
whatever I got divided by mass of the
airing of bond in 200 and you saw all
they triggered something else thirty
four point zero percent easy early
let's take one more numerical similar
thing but instead of bromine I have
chlorine here so I have my all any
component right it has some chlorine so
it a is b AG CL so mass of chlorine is
what 35.5 mass of EDC it was one zero
eight plus thirty five point five one
four do you divide by right the AG
series
silver is one zero eight plus thirty
five point five correct so if you solve
this you get one forty seven five so one
43.5 AG CL if you odd that means you
have thirty five point five chlorine
right
so if I have got one gram of silver
chloride this implies I have thirty five
point five by one forty three point five
gram of pure Union or any compound
number for how much I have got I have
got this one gram of sugar pure I swear
by 0.7 by by five seven four zero gram
of AG CL and this implies that I have
thirty five point five one for the three
point I will do 0.57 for 0 gram of
fluorine in Hagen example so that this
one's grams of chlorine organic compound
my percentage of chlorine is what this
one's gram what I have got divided by
total mass of avoiding a bond that is
this guy in two hundred so you solve
this you get thirty seven point five six
but someplace that's correct
thank you visit exam viacom to watch
free education videos try free on my
debts get the best quality study
materials study from the best tutors and
mentors and much more thanks again