Discussion
If xA and xB are the respective mole fractions of A and B in an ideal solution of the two and TA, TB, T are the fusion temperatures of pure A, pure B and the ideal solution respectively, then
Answer: D
The correct relation is the one with A's enthalpy of fusion, A's melting point and a negative exponent: \(1-x_{B}=\)\(\exp[\frac{-\Delta H^{o}_{fus(A)}}{R}(\frac{1}{T}-\frac{1}{T_{A}})]\).
The solution freezes when its component A is in equilibrium with pure solid A, so the condition involves A's enthalpy of fusion and A's melting point. For an ideal solution:
\[ \begin{aligned} \ln x_{A} &= -\frac{\Delta H_{fus(A)}}{R}\left(\frac{1}{T} - \frac{1}{T_{A}}\right) \\ x_{A} &= 1 - x_{B} \end{aligned} \]
Exponentiating gives the last option. A check: T lies below TA, so 1/T - 1/TA is positive, the exponent is negative and xA is less than 1, as it must be.
The two options with B's enthalpy and TB describe solid B separating instead, and the option with a positive exponent would make xA greater than 1.
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