To solve this problem, we first need to understand how the depression in freezing point and the elevation in boiling point can be related through their respective formulas involving the molalities of the solution.
The thermodynamic equations related to the colligative properties for non-dissociative and nonassociate compounds are:
The depression in freezing point formula is given by:
∆Tf=Kf⋅mwhere
∆Tf is the depression in freezing point,
Kf is the cryoscopic constant (freezing point depression constant), and
m is the molality of the solution.
The elevation in boiling point formula is given by:
∆Tb=Kb⋅m where
∆Tb is the elevation in boiling point,
Kb is the ebullioscopic constant (boiling point elevation constant), and
m remains the molality of the solution.
Given that the depression in freezing point is
aK, and using our first equation:
a=Kfâ‹…mWe can solve for
m (molality):
m=‌.Now inserting this value of molality into the equation for the elevation in boiling point:
∆Tb=Kb⋅‌ Given that the boiling point elevation constant
Kb is provided, and no transformation is applied to the constant itself, the relationship can be set up directly to find
∆Tb, which in the context of the problem can be considered as
X. Thus, the formula for
X becomes:
X=Kb⋅‌Here,
Kb=cK‌kg‌mol−1,Kf=bK‌kg‌mol−1, and the depression in the freezing point is given by
a. Thus substitution gives:
X=c⋅‌. Comparing this result with the options provided:
Option A,
X=2c×‌, (does not match)
Option B,
X=c×‌, (does not match)
Option C,
X=c×‌, (matches)
Option
D,X=c×‌, (does not match)
So the correct answer is Option C:
X=c×‌.