Suppose we have several samples that have unknown tin and bismuth compositions a

Question

Suppose we have several samples that have unknown tin and bismuth compositions and we measure their cooling curves. Each sample has a total mass of 10 grams. We assign a series of letters to each of the samples ranging from A to F. In lab, we measure the following values for T1 and T2, i.e. the first (liquidus) and second phase (solidus) transformation temperatures respectively Table 1: The phase transformation temperatures for the six Sn-Bi samples. T1 (degrees C) 271.4 139.5 180.0 165.1 165.0 200.8 T2 (degrees C) none none 139.4 139.6 139.4 139.3 Sample l intentionally conceal the weight percent of bismuth for each of the compositions shown in Table 1 above but provide you with the following relationships. D-C+8wt. % Bi. (1) E=F-10wt%Bi. (2) Based on this information and using the reference diagram in Figure 1 as a guide, fill out the following table showing the approximate weight percent of each sample. I suggest you print out this assignment and use a ruler. Each correct weight percent composition (to within 1%) is worth 3 points. You get 2 additional points for legible work shown in the diagram for a total of 20 points. Note that there is a difference between weight and atomic percent! Table 2: Compositions of the six unknown samples Sample Weight percent Bi

Explanation / Answer

Let us first analyse the information in the given table in comparison with the Bi- Sn phase diagram. Each sample is a point of this diagram.

The sample A has T1 at 271.4°C which can be approximated to the 271°C melting point of PURE Bi read from the phase diagram also T2 is absent which is in agreement with the Bi+L solid-liquid biphase present below T1.

The sample B has T1 at 139.5°C which is the approximate eutectic temperature ,Therefore B is on the horizontal eutectic temperature line. Absence of T2 means that it has both the phase changes at a unique point which by definition is the EUTECTIC POINT.

Now when it comes to the points C and D, the relation is that percentage of Bi in D is 8% more than that present in sample C. And since the T1 of C =180.0°C is more than T1 of D = 165.1°C which is an indication of negative slope, both the points are to the left hand side of Eutectic point. The liquidus line though it looks like a curve, we can take it as a line for our calculations because so equation of curve was given. {Note: It is better to neglect the minute difference in T2 for easing the math}

Bi% at 232°C on the liquidus line is 0%.The line eventually ends at Eutectic point which is at 139°C and contains 57% Bi. Let's use these key points to find C and D with the help of the slope of liquidus line.

Finding C:
(232-180) /(232-139) =(0%-c%) /(0%-57%)
Which gives C as containing 31.87%(approx.~32%) of Bi

Similarly D can be derived from
(232-165.1) /(232-139) =(0%-d%) /(0%-57%)
Which gives D as containing ~41% of Bi

Now E and F are related such that the percentag of Bi in E is 10% less than present in F. And since the T1 of E=165.0 is less than T1 of F=200.8 which indicates a line with a positive slope. Both the points are to the right hand side of Eutectic point. {Note: It is better to neglect the minute difference in T2 for easing the math}

So a similar logic can be used to find E and F to that of C and D.
Bi% at 139°C on the liquidus line is 57%.The line ends at Pure Bi (100%) which is at 271°C.

Finding E:
(139-165) /(139-271) =(57%-e%) /(57%-100%)
Which gives E as containing 65.47%(approximately~65.5%) of Bi

For F,
(139-200.8) /(139-271) =(57%-f%) /(57%-100%)
Which gives F as containing 77.13% of Bi

ANSWER:   { (Sample ;   weight%(Bi)) }=
{ (A,100%), (B,57%), (C,32%), (D,41%), (E,65.5%), (F,77%) }

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