JEE Main & Advanced Chemistry States of Matter Question Bank Kinetic molecular theory of gases and Molecular collisions

  • 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 \[c{{m}^{3}}\]

    Correct Answer: C

    Solution :

               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 \[\Rightarrow 0.6\,c{{m}^{3}}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner