Amixture of helium and nitrogen gases at a total pressure of 641 mm mm hg. if the gas mixture contains 0.399 grams of helium, how many grams of nitrogen are present? hg contains helium at a partial pressure of 231 g

Answer : The mass of nitrogen present are, 4.965 grams.

Explanation :

According to the Raoult's law,

where,

= partial pressure of gas = 231 mmHg

= total pressure of gas = 641 mmHg

= mole fraction of helium gas = ?

Now put all the given values in this formula, we get the mole fraction of helium gas.

Now we have to calculate the mole fraction of nitrogen gas.

Now we have to calculate the mass nitrogen gas.

where,

n = moles, w = mass, M = molar mass

Now put all the given values in this expression, we get:

Therefore, the mass of nitrogen present are, 4.965 grams.

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

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