[baboom]

I have a sample of salt, which contains only KCl and NaCl. The total mass of the salt is 1g. The total number of moles of chloride in the sample is 1.5e-3 moles.

Is it possible to calculate the mole fraction of KCl and also the mole fraction of NaCl in the sample?

[omkar]

let x be mole fraction of NaCl& y for  KCl
From given data
          x(mwt of NaCl)+y(molecular wt of KCl)=1g
           x+y=1.5e-3  (SINCE BOTH MOLECULES CONTAINS Cl ATOMS IN 1:1 RATIO)

[baboom]

Thanks for your reply.
I have tried that method, but my mole fraction comes out to be a negative number. Do you know what I may be doing wrong:

Let x=mole fraction of NaCl=nNaCl/nCl
and y=mole fraction of KCl=nKCl/nCl
Therefore, x+y=1

Let M=molecular Weight of NaCl and N=molecular weight of KCl

M*x+N*y=mass of sample/nCl; since x+y=1,
M*x+N*(1-x)=mass of sample/nCl=M*x+N-N*x
mass of sample/nCl=x(M-N)+N
(mass of sample/nCl)-N=x(M-N)

[(mass of sample/nCl)-N]/(M-N)=x This is the final equation I got to calculate the mole fraction of NaCl.

The problem is that the denominator is a negative number (since M<N) and the numerator is a positive number. This gives me a negative answer, which is wrong. I am not sure what I did wrong here, since all the steps I followed look very logical to me.

[Borek]

M*x+N*y=mass of sample/nCl

Units don't match. omkar gave you the correct equation, although x&y were not mole fractions, just numbers of moles of both chlorides.

[zhyx00]

I'm not quite understand what does "mole fraction" mean. but if "omkar" 's answer is right ,then I suppose you make a mistake:"Let x=mole fraction of NaCl=nNaCl/nCl
and y=mole fraction of KCl=nKCl/nCl
Therefore, x+y=1". I think it should be x+y=1.5e-3 , not 1.
another point , if I have got the exact meaning of this question that means that the total weight and total number of moles of the mix with two componets are both given and the molecular weight of KCL and NaCL is known(74.55 and 58.41 respectively), then the answer can be decided without calculation. reason: there are only 2 numbers uncertain actually, x and y , and correspondingly 2 equations can be built x +y =1.5e-3 and 74.55x + 58.41y=1, so we have 2 uncertain numbers and 2 equations , this 2 nembers can be calculated definitely. we call it a group of duality simple eqation. this method of judgement before calculation is more important than this question, I think. I hope my answer will be useful to you.

[Borek]

then the answer can be decided without calculation.

As you have later explained how to calculate the answer, I think you mean here "existence of the answer can be decided without calculation". But that's not necesarilly true, it may happen that equations are not independent or contradictory.

[zhyx00]

yes.I agree with you. you are precise.

[baboom]

Thanks for your reply guys.

I decided to use a different method, although I am not sure if it is correct:

Let M=molecular weight:

M(Nacl)/(M(KCl)+M(NaCl))
=mole fraction of KCl

[Borek]

No, that's wrong. Using this method you will always get identical answer no matter what the mixture composition is.

Use just a definition of molar fraction, you don't have to invent it.

[baboom]

 thanks Borek, that site is very helpful