• # question_answer At ${{100}^{o}}C$ and 1 atm, if the density of liquid water is 1.0 g $c{{m}^{-3}}$ and that of water vapour is 0.0006 g ${{m}^{-3}}$, then the volume occupied by water molecules in 1 litre of steam at that temperature is  [IIT 2000] A)                 6 $c{{m}^{3}}$                B)                 60 $c{{m}^{3}}$ C)                 0.6 $c{{m}^{3}}$             D)                 0.06 $c{{m}^{3}}$

Volume of steam = 1lt = ${{10}^{3}}c{{m}^{3}}$                    $\because m=d.V$                    \mass of ${{10}^{3}}c{{m}^{3}}$steam = density ´ Volume                    = $\frac{0.0006gm}{c{{m}^{3}}}\times {{10}^{3}}c{{m}^{3}}$ $=0.6gm$                    Actual volume occupied by ${{H}_{2}}O$ molecules is equal to volume of water of same mass                    $\therefore$ Actual volume of ${{H}_{2}}O$ molecules in $6gm$ steam                    = mass of steam/density of water                                 = $0.6\,gm$/1 gm/cm3 $\Rightarrow 0.6\,c{{m}^{3}}$
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