Question

The volume occupied by $3.5$g of $O_2$ gas at ${27}^o$C & $740$ mm of Hg pressure$[O=16]$

Answer 1

Solution:-

Mass of $O_2=3.5g$

Molar mass of $O_2=32g$

$\therefore$ No. of moles of $O_2,\left(n\right)=\frac{3.5}{32}=0.109\text{mol}$

Given that:-

Pressure, $\left(P\right)=740\text{mm of Hg}=0.97atm$

Temperature, $\left(T\right)=27℃=(27+273)=300K$

$R=0.0821\text{L atm}{K}^{-1}{mol}^{-1}$

Now from ideal gas law,

$PV=nRT$

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

$\Rightarrow V=\frac{0.109\times 0.0821\times 300}{0.97}=2.76L$

Hence the required answer is $2.76L$.