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tangare [24]
2 years ago
11

Melts in the system pb-sn exhibit regular solution behavior. at 473°c apb = 0.055 in a liquid solution of xpb = 0.1. calculate t

he value of w for the system and calculate the activity of sn in the liquid solution of xsn = 0.5 at 500°c
Chemistry
1 answer:
Masteriza [31]2 years ago
6 0
Given the temperature 746 K and activity of Pb equal to 0.055. The mole fraction of Pb is 0.1. So, the mole fraction of Sn = 0.9.Activity coefficient, γ = 0.055 / 0.1 = 0.55.The expression for w=ln⁡〖γ_Pb x RT〗/(X_Sn^2 )=(-0.5978 x 8.314 J/(mol K ) x 746 K)/(0.9 x 0.9)= -4577.7 J= -4578 J

Now we use the computed value above and new temperature 773 K. The mole fraction of Sn and Pb are 0.5 and 0.5 respectively. Calculate the activity coefficient in the following manner.lnγ_Sn=w/RT  X_Pb^2=(-4578 J)/(8.314 J/mol  x 773 K)  x 0.5 x 0.5= -0.718lnγ_Sn=exp⁡(-0.178)=0.386The activity of  Sn= γ_Sn  x X_Sn=0.386 x 0.5=0.418
w of the system is -4578 J and the activity of Sn in the liquid solution  of xsn at 500 degree Celsius is 0.418
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Answer:

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Explanation:

First of all we need to know, that a weight percent represents, the mass of solute in 100 g of solution.

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Density always refers to solution → Solution density = Solution mass / Solution volume

1.67 g/mL = 100 g / Solution volume

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To determine molarity (mol/L) we must convert the mL to L

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8 0
2 years ago
A mixture of three gases has a pressure at 298 K of 1380 mm Hg. The mixture is analysed and is found to contain 1.27 mol CO2, 3.
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Answer:

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)

Explanation:

<u>Step 1:</u> Data given

A mixture of three gases has a total pressure of 1380 mm Hg (=1.81579 atm) at 298 K

Moles of CO2 = 1.27 moles

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Total number of moles = n(CO2)+ n(CO)+ n(Ar) = 1.27 mol+ 3.04 mol+ 1.50 mol = 5.81 moles

<u>Step 3:</u> Calculate mol fraction Ar

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<u>Step 4</u>: Calculate partial pressure

1380 mm Hg * 0.258 moles Ar = 356.04 mm Hg = 0.4685 atm

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)

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5 0
2 years ago
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Answer:

See explanation

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