Avolume of 18.0 l contains a mixture of 0.250 mole  n 2 , 0.250 mole o 2 , and an unknown quantity of he . the temperature of the mixture is 0 ∘ c , and the total pressure is 1.00 atm . how many grams of helium are present in the gas mixture?

Answer : The mass of helium present in the gas mixture are, 1.216 grams.

Solution : Given,

Moles of = 0.25 mole

Moles of = 0.25 mole

Total volume of gas mixture = 18 L

Total pressure of gas mixture = 1 atm

Total temperature of gas mixture =

Molar mass of helium = 4 g/mole

First we have to calculate the total moles of mixture of gas.

using ideal gas equation,

where,

P = total pressure of gas mixture

T = total temperature of gas mixture

n = total mole of gas mixture

V = total volume of gas mixture

R = gas constant = 0.0821 Latm/moleK

Now put all the given values in the above formula, we get the total moles of mixture of gas.

Now we have to calculate the moles of helium gas.

Now we have to calculate the mass of helium.

Therefore, the mass of helium present in the gas mixture are, 1.216 grams.

The mass of helium gas that present in the gas mixture is 1.216 grams.

The ideal gas equation

where,

P = total pressure of gas mixture

T = total temperature of gas mixture

n = total mole of gas mixture

V = total volume of gas mixture

R = gas constant = 0.0821 Latm/moleK

Put the values, and solve it for n,

The number of moles of the helium gas

The mass of the helium gas

Therefore, the mass of helium gas that present in the gas mixture is 1.216 grams.

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the answer is na (sodium)