Question

Diagram 3 Solid Liquid Phase Diagram For Organics A And B 80 60 Temp °c) 40 20 0 0 0.2 0.6 0.8 1.0 Mole Fraction of Organic M
Refer to the Diagram 3 The solid-liquid phase diagram for mixtures of Organic A (molar mass: 115.5 g/mole) and Organic B (mol
Refer to the Diagram 3 The solid-liquid phase diagram for mixtures of Organic A (molar mass: 115.5 g/mole) and Organic B (mol
Refer to the Diagram 3 The solid-liquid phase diagram for mixtures of Organic A (molar mass: 115.5 g/mole) and Organic B (mol
Refer to the Diagram 3 The solid-liquid phase diagram for mixtures of Organic A (molar mass: 115.5 g/mole) and Organic B (mol
Refer to the Diagram 3 The solid-liquid phase diagram for mixtures of Organic A (molar mass: 115.5 g/mole) and Organic B (mol
Refer to the Diagram 3 The solid-liquid phase diagram for mixtures of Organic A (molar mass: 115.5 g/mole) and Organic B (mol
Refer to the Diagram 3 The solid-liquid phase diagram for mixtures of Organic A (molar mass: 115.5 g/mole) and Organic B (mol
Refer to the Diagram 3 The solid-liquid phase diagram for mixtures of Organic A (molar mass: 115.5 g/mole) and Organic B (mol
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Answer #1

Answer

First question.      The temperature at which pure B melts, Tf = 58.2 °C

Second question. Option d. A and B but not eutectic

Third question.     The mole fraction of A remaining in the Liquid phase is "zero", as the substance will have no liquid phase.

Fourth question.   Temperature of melting was 30.0 °C

Fifth question. Eutectuc mixture melts as 20.0 °C

Sixth question.      The mixture is melting at 72.0 °C

Seventh question. Mole fraction of A in Eutectuc mixture is 0.24

Eighth question.    Pure A melts as 80.0 °C

Explanation

First question.      refer from diagram 3 (temperature in red)

Second question:

nA = 1.5 mole ; nB = 7.25 mole

mole fraction of A = nA/(nA+nB)

                             = 1.5/(1.5+7.25)

                             = 1.5/8.75

                             = 0.1714

By interpreting in the graph along the mole fraction 0.1714, the answer is A and B but not eutectic.

Third question. Refer to green line in diagram 3

Fourth question

20.0 g of A and 39.7 g of B

nA = 20/115.5 = 0.1732

nB = 39.7/98.4 = 0.4035

mole fraction of A = 0.1732/(0.1732+0.4035)

                             = 0.300

When interpreted this value of mole fraction of A in the diagram 3, the temperature melting was 30.0 °C

Fifth question

Eutectic mixture melts at eutectic point

Sixth question

50 g of A 20 g of B

nA = 50/115.5 = 0.433

nB = 20/98.4 = 0.2033

mole fraction of A = nA/(nA+nB)

                             = 0.433/(0.433+0.2033)

                             = 0.433/0.6363

                             = 0.6805

The mixture is melting at 72 °C

Seventh question

Mole fraction of A in Eutectuc mixture is 0.24

Eighth question.

Refer to value in pink in diagram 3

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