Question

A $34.0$ litre contains contains $212g$ of $O_2\left(g\right)$ at $27{}^{o}C$. What mass of $O_2\left(g\right)$ must be released to reduced the pressure to $2.463$ atm?

• A. 112.5 g
• B. 103.2 g
• C. 120.5 g
• D. 125.6 g

Answer 1

Answer: (B)

103.2 g

Number of moles of oxygen present = $\frac{212}{32}=6.625$

$Since,PV=nRTSo,P=\frac{nRT}{V}P=\frac{6.625\times 0.08206\times 300}{34}P=4.8atm$.

Now for new conditions, P = 2.463 atm.

$Here,V=constant,P_1=4.8atm,P_2=2.463atm{n}_1=6.625So,\frac{P_1}{P_2}=\frac{n_1}{n_2}\frac{4.8}{2.463}=\frac{6.625}{n_2}{n}_2=3.4$

So, number of moles needs to released = 6.625 - 3.4 = 3.225

So, mass of oxygen released = $3.225\times 32=103.2g$.

So, correct answer is option B.