A solution consisting of isobutyl bromide and isobutyl chloride is found to have a
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
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To determine the molar composition of the mixture, we need to use the linear relation between refractive index and the molar composition. This relation assumes that the refractive indices of the pure components can be expressed as a linear combination of the refractive indices of the mixture.
Let's use the following variables:
- x: molar fraction of isobutyl bromide (IBB)
- y: molar fraction of isobutyl chloride (IBC)
First, let's express the refractive index of the mixture (RM) as a linear combination of the refractive indices of the pure components:
RM = x * RIBB + y * RIBC
Substituting the given refractive indices at 20oC:
1.3931 = x * 1.4368 + y * 1.3785
Now, we have one equation with two unknowns (x and y). To solve it, we need another equation. Since x and y represent molar fractions, their sum should be equal to 1:
x + y = 1
To solve these simultaneous equations, we can use substitution or elimination method. Let's use the elimination method in this case.
Rearrange the second equation to obtain x = (1 - y).
Substitute x in the first equation:
1.3931 = (1 - y) * 1.4368 + y * 1.3785
Now solve for y:
1.3931 = 1.4368 - 1.4368y + 1.3785y
Combine like terms:
1.3931 = 1.4368 - 0.0583y
Rearrange the equation to isolate y:
0.0583y = 1.4368 - 1.3931
0.0583y = 0.0437
Divide both sides by 0.0583:
y = 0.0437 / 0.0583
y ≈ 0.7493
Substitute this value of y back into the equation x + y = 1 to find x:
x + 0.7493 = 1
x = 1 - 0.7493
x ≈ 0.2507
The molar fraction (in percent) of isobutyl bromide (IBB) is obtained by multiplying x by 100:
Molar fraction of IBB = x * 100
Molar fraction of IBB ≈ 0.2507 * 100
Molar fraction of IBB ≈ 25.07%
Therefore, the molar composition (in percent) of the mixture is approximately 25.07% isobutyl bromide and 74.93% isobutyl chloride.