Question

The diagram given below shows phase relations between components P and Q at 1 bar pressure. If 'X' represents the initial liquid composition, which of the following statements is/are CORRECT during equilibrium crystallization?

• A. Initial liquid composition is 60 wt.% of P and 40 wt.% of Q
• B. The composition of the solid in equilibrium with the liquid at ‘Y' is 10 wt.% of P and 90 wt.% of Q
• C. The bulk composition of the final solid product is 40 wt.% of P and 60 wt.% of Q
• D. The proportion (on the basis of wt.%) of two phases, Mss: Nss is 1:2 at 750 °C

Hint:

In binary diagrams:
- Bulk composition is conserved (equilibrium crystallization).
- The composition of coexisting phases at a given \(T\) is read from tie–line ends.
- Phase proportions at that \(T\) come from the lever rule: lengths opposite each phase are proportional to its fraction.

Answer 1

The Correct Option is B, C, D

Step 1: Read the bulk (initial) composition from point X.
Point \(X\) lies at \(\approx 60\) wt.% \(Q\) (hence \(\approx 40\) wt.% \(P\)). Therefore option (A) stating “40 wt.% \(Q\)” is false.
Step 2: Solid in equilibrium with the liquid at \(Y\).
At temperature \(Y\) (on the liquidus), draw the tie–line to the solidus on the right-hand side. The intercept on the solidus lies in the N\textsubscript{ss} field at \(\approx 90\) wt.% \(Q\) (\(\approx 10\) wt.% \(P\)). Hence the solid coexisting with the liquid at \(Y\) is \(\sim\)N\textsubscript{ss}(10\(P\)–90\(Q\)), so (B) is correct.
Step 3: Bulk composition after equilibrium crystallization.
Under \emph{equilibrium} (batch) crystallization, the bulk composition of the \emph{final aggregate of solids} must equal the initial bulk composition of the system. Since \(X\) is \(\approx 40\,P\)–\(60\,Q\), the final solid aggregate is also \(40\) wt.% \(P\) and \(60\) wt.% \(Q\). Thus (C) is correct.
Step 4: Proportions of M\textsubscript{ss and N\textsubscript{ss} at 750\,$^\circ$C (lever rule).}
At \(750\,^\circ\mathrm{C}\) the horizontal tie–line intersects the solvus at \(\approx 40\) wt.% \(Q\) (M\textsubscript{ss}) and \(\approx 70\) wt.% \(Q\) (N\textsubscript{ss}). For the bulk \(C_0 = 60\) wt.% \(Q\): \[ \text{fraction of M}_{ss}=\frac{C_N-C_0}{C_N-C_M}=\frac{70-60}{70-40}=\frac{10}{30}=\frac{1}{3},\quad \text{fraction of N}_{ss}=\frac{C_0-C_M}{C_N-C_M}=\frac{60-40}{70-40}=\frac{20}{30}=\frac{2}{3}. \] Hence \( \text{M}_{ss} : \text{N}_{ss} = 1:2\). Therefore (D) is correct.
\[ \boxed{\text{Correct statements: (B), (C), and (D).}} \]