Question

A mixture 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 1

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

Explanation :

According to the Raoult's law,

`p_(He)=X_(He)* p_T`

where,

`p_(He)` = partial pressure of gas = 231 mmHg

`p_T` = total pressure of gas = 641 mmHg

`X_(He)` = mole fraction of helium gas = ?

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

`231mmHg=X_(He)* 641mmHg`

`X_(He)=0.36`

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

`X_(He)+X_(N_2)=1`

`X_(N_2)=1=0.36=0.64`

Now we have to calculate the mass nitrogen gas.

`(X_(He))/(X_(N_2))=(n_(He))/(n_(N_2))`

`(X_(He))/(X_(N_2))=((w_(He))/(M_(He)))/((w_(N_2))/(M_(N_2)))`

where,

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

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

`(0.36)/(0.64)=((0.399)/(4))/((w_(N_2))/(28))`

`w_(N_2)=4.965g`

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