1plz help
What maximum volume of 3M KOH can be prepared from 1L each of 1M KOH and 6M KOH ....
(1) By using water
(2) Without using water
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6 Answers
11(1) for Ist one :
add all the amount of 1M and 6M solution
you would get 2 L, 3.5M solution of KOH
now you have to make its molarity = 3M by adding water
using the formula of dilution:
M1V1 = M2V2
(where M1=3.5 ,M2=3 ,V1 = 2L)
you'll get the value of V2=2.33L which is the maximum volume in first case
(2) for IInd one :
let us mix V1 of Ist solution and V2 of second
M1V1 + M2V2 = M3V3
ie. V1 + 6V2 = 3(V1 + V2)
from this you get,
V1V2 = 32------------------eq(1)
we have to maximise :V1 + V2
it is clear that using full volume of one of the solution and finding the volume of other using eq(1) will get us the answer
lets use V2 full (ie 1L) then V1 = 1.5L but we don't have this amount of V1
so we will use full amount of V1 (ie 1L again) and then V2=0.67L (which is available)
so maximum amount of volume obtainable in second part :
V1 + V2 = 1.67L
1357For second one
1 L of 1M and y L of 6M
so we get 1*1+6y=3(1+y)
or 3y=2
y=2/3
so total volume= 1+2/3 L
1357For first one
As we have to find maximum volume.
So maximum is mole of KOH maximum is the volume
So we mix both the solution completely and then add water