Question

A gaseous mixture containing $0.35g$ of $N_2$ and 5600 ml of $O_2$ at STP is kept in a 5 litres flask at 300K. The total pressure of the gaseous mixture is :

• A. 1.293 atm
• B. 1.2315 atm
• C. 12.315 atm
• D. 0.616 atm

Answer 1

Answer: (A)

1.293 atm

As we know that,

$\text{No. of moles}=\frac{\text{mass}}{\text{molar mass}}$

Mass of $N_2=0.35g$

Molar mass of $N_2=28g$

No. of moles of $N_2=\frac{0.35}{28}=0.0125\text{mol}$

Volume of $O_2=5600mL$

Volume at NTP $=22400mL$

No. of moles of $O_2=\frac{5600}{22400}=0.25\text{mol}$

Given that :

Volume of flask $\left(V\right)=5L$

Temperature of flask $\left(T\right)=300K$

Now from ideal gas law :

$PV=nRT$

$\Rightarrow P=\frac{nRT}{V}$

Therefore,

Partial pressure of $O_2,\left(P_{O_2}\right)=\frac{0.25\times 0.0821\times 300}{5}=1.2315atm$

Partial pressure of $N2,\left(P_{N_2}\right)=\frac{0.0125\times 0.0821\times 300}{5}=0.0615atm$

$\therefore$ Total pressure, $\left(P\right)=P_{O_2}+P_{N_2}=1.2315+0.0615=1.293atm$

The correct option is A.