11q2.....
we know @ = √(Kac)
so @1@2 = √(Ka1Ka2)
substituing
@1@2 = √9 x 10-510-5 = 3
@1 = 3 @2
where @1 = degree of dissociation of HA
@2 = degree of dissociation of HB
so HA is 3 times stronger than HB...
(1)A certain gas is hexamerised 2 a small extent.If initially 1 mole gas is taken the ratio \frac{PV}{RT} will b equal to
(A)1-\frac{5K_{c}}{V^{5}} (B)\frac{5K_{c}}{V^{5}} (C)1-\frac{5K^{2}_{c}}{V^{5}} (D)1-\frac{4K^{2}_{c}}{V^{4}}
(2)If dissociation constants of two acids A and B are 9\times 10^{-5} and 10^{-5} respectively. Acid HA is how much time stronger than HB?
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5 Answers
1option (a)
but the q itself is not very right you have to give me the eq or the dimensions of the Kc it can be solved by looking on the options but yet it is not the right way
111Sorry,Cant understand the first one.
This much is given in the question,So please explain.
1357It looks to be (a) here
6A=A6
1 0
1-x x/6
if x is very very small then we can say that 1-x=1
Kc=(x/6V)/(1/V)6 or x=6Kc/V5
PV/RT=total number of moles
=1-x+x/6=1-5x/6=1-5Kc/V5
1hmm that's how it's done thnks Manish bhaiya ....... thnks a lot .... mujhe to wo Kc kahan se aaya samjh main hi nahin aa raha tha........
ANY WAYS thnks again [1][1]