Question

$16$gm of oxygen and $3$ gm of hydrogen are present in a vessel at $0^0$C and $760$mm of Hg pressure. Volume of the vessel is:

• A. $22.4$L
• B. $44.8$L
• C. $11.2$L
• D. $5.6$L

Answer 1

Answer: (B)

$44.8$L

Solution:- (B) $44.8L$

Mol. wt. of $H_2=2gm$

Mol. wt. of $O_2=32gm$

As we know that,

No. of moles $=\frac{\text{Wt.}}{\text{Mol. wt.}}$

Wt. of $O_2$ in the vessel $=16gm$

Wt. of $H_2$ in the vessel $=3gm$

Therefore,

No. of moles of $H_2$ in the vessel $=\frac{3}{2}=1.5\text{mole}$

No. of moles of $O_2$ in the vessel $=\frac{16}{32}=0.5\text{mole}$

$\therefore$ Total no. of moles in the vessel $\left(n\right)=1.5+0.5=2\text{moles}$

Given:-

Pressure in the vessel $\left(P\right)=760\text{mm of Hg}=1atm$

Temperature $\left(T\right)=0℃=273K$

Now, from ideal gas equation, we have

$PV=nRT$

$1\times V=2\times 0.0821\times 273$

$\Rightarrow V=44.8276L$

Hence the volume of the vessel is $44.8L$.