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Kryger [21]
3 years ago
12

n a system for separating gases a tank containing a mixture of hydrogen and carbon dioxide is connected to a much larger tank wh

ere the pressure is kept very low. The two tanks are separated by a porous membrane through which the molecules must effuse. If the initial partial pressures of each gas is 5.00 atm, what will be the mole fraction of hydrogen in the tank after the partial pressure of carbon dioxide has declined to 4.50 atm
Chemistry
1 answer:
Stella [2.4K]3 years ago
4 0

Answer:

The mol fraction of the hydrogen gas = 0.526

Explanation:

Step 1: data given

mixture of hydrogen and carbon dioxide

the initial partial pressures of each gas is 5.00 atm

the partial pressure of carbon dioxide has declined to 4.50 atm

Step 2: Calculate total pressure

Total pressure = 4.5 atm + 5.0 atm

Total pressure = 9.5 atm

Step 3: Calculate mol fraction

Partial pressure = mole fraction * total pressure

Mol fraction = partial pressure / total pressure

Mol fraction carbon dioxide = 4.5 atm / 9.5 atm

Mol fraction carbon dioxide =  0.474

Mol fraction of hydrogen = 1 - 0.474 = 0.526

The mol fraction of the hydrogen gas = 0.526

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It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

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Where : ΔTf = change in freezing point (°C)

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m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

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ΔTf = 1.69 ° C

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