Question

\( 2.8 g \) of \( N _{2}, 2.8 g CO , 4.4 g CO _{2} \) are found to exert a pressure of 700 Torr. Find partial pressure of \( N _{2} \) gas in the mixture - (A) \( 280.8 \) Torr (B) \( 233.3 \) Torr (C) 300 Torr (D) None of these

Answer 1

We have — given,

weight of N₂ = 2.8 g

molecular weight of N₂ = 28 g/mol

Number of moles of N₂ \(=\frac{2.8\,g}{28\,g/mol}\)

= 0.1 mol

weight of CO = 2.8 g

molecular weight of CO = 28 g/mol

Number of moles of CO \(=\frac{2.8\,g}{28\,g/mol}\)

= 0.1 mol

weight of CO₂ = 4.4 g

molecular weight of CO₂ = 44 g/mol

Number of moles CO₂ \(=\frac{4.4\,g}{44\,g/mol}\)

= 0.1 mol

Mole fraction of N₂ \(=\frac{\text{moles of }N_2\,gas}{\text{Total moles of gas}}\)

= 0.1/0.1+0.1+0.1

= 0.1/0.3

= 0.33333.....

We know that —

P_A = x_AP

where, P_A = Partially pressure of 'A'

x_A = mole fraction of 'A'

P = Total pressure = 700 Torr.

Therefore,

Partially pressure of N₂ = 0.3333 x 700 Torr

= 233.3 Torr

Hence option (B) is right answer.