A solution consisting of isobutyl bromide and isobutyl chloride is found to have a

Question

refractive index of 1.3931 at 20oC. The refractive indices at 20oC of isobutyl bromide and isobutyl chloride are 1.4368 and 1.3785, respectively. Determine the molar composition (in percent) of the mixture by assuming a linear relation between the refractive index and the molar composition of the mixture.

How do I find this assuming that the total is 100%?
PLEASE EXPLAIN!
THANK YOU!

Answer 1

Let's call mole fraction isobutylchloride = C.
and mole fraction isobutylbroomide = B

Now if we draw a straight line from left to right
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and label the left index as 0 mole fraction B (meaning that is pure C and no B) and the right index as 1.0 mole fraction B (meaning pure B and no C).
Now the left index is marked as 1.3789 refractive index (pure C) and the right index is marked as 1.4368 (pure B). Check me out on that.
What we want to do is to find the fraction that 1.3931 represents. The difference from left to right is
1.4368-1.3785- ??.
The difference from 1.3931(the sample) and the left mark is -1.3785 = ??.
Now take the ratio of the two and that should be the mole fraction of B in the mixture. That times 100 should convert to percent. Check my thinking.

Answer 2

Thank you!

Answer 3

What is the percent by mass of Ca3N2?