Question

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

• A. 1L
• B. 5.8L
• C. 2.27L
• D. 5.6L

Answer 1

The correct option is C 2.27L

PV $=$ nRT

$V=nRT$ $÷$P

Mole $=$ Given mass $÷$ Molar mass

$=3.5$ $÷$ 32 = 0.11 mol

​R = 0.0821 L atm $K^{-1}mo{l}^{-1}$

T = 300K

​P = 740 mm Hg = 0.973 atm (760 mm Hg = 1atm)

V = 0.11 x 0.0821 L atm $K^{-1}mo{l}^{-1}$ x 300K $÷$ 0.973 atm

$=2.77L$