Question
The volume occupied by $$88\,g$$ of $$C{O_2}$$ at $$30{\,^ \circ }C$$ and $$1\,bar$$ pressure will be
A.
5.05$$\,L$$
B.
50.36$$\,L$$
C.
2$$\,L$$
D.
55$$\,L$$
Answer :
50.36$$\,L$$
Solution :
$$m = 88\,g,M = 44\,\,g\,\,mo{l^{ - 1}},$$ $$T = 273 + 30 = 303\,K$$
$$\eqalign{
& P = 1\,bar,V = ? \cr
& PV = nRT\,\,{\text{or}}\,\,V = \frac{{nRT}}{P} = \frac{m}{M}\frac{{RT}}{P} \cr
& V = \frac{{88 \times 0.0831 \times 303}}{{44 \times 1}} = 50.36\,L \cr} $$