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