112ml. of oxygen at STP is subjected to liquefication. The mass of liquid oxygen obtained is

One mole of gas occupies 22400 mL at STP conditions, so the mole carry by 112 mL is:

$\mathrm{Moles}=\frac{112 \mathrm{mL}}{22400 \mathrm{mL}}\times 1 \mathrm{mole}=0.005 \mathrm{mole}$

Weight of oxygen present in 0.005 mole is calculated as:

$\mathrm{Weight}=0.005 \mathrm{mole}\times 32 \mathrm{g}/\mathrm{mol}=0.16 \mathrm{g}$=