Grams of butane (C4H10) formed by the liquefaction of 448 litres of the gas (measured at (STP) would be

Answer: Option C

At S.T.P 1 mole of any gas occupies $$22.4ltrs$$
Given $$448ltrs$$   so, number of moles = $$\frac{448}{22.4}$$ $$=20$$
We know number of moles = $$\frac{\text{weight}}{\text{gram moleculear weight}}$$     $$\left(\text{molecular weight of }{C}_4{H}_{10}=58\right)$$
So, $$20=$$ $$\frac{\text{weight}}{58}$$
Hence weight of $$C_4{H}_{10}=1160$$