Saved Bookmarks
| 1. |
The mass of one litre sample of ozonised oxygen at N.T.P. was found to be 1.5 g. When 100 mL of this mixture at N.T.P. were treated with turpentine oil, the volume was reduced to 90 mL. Calculate the molecular mass of ozone. |
|
Answer» Solution :As ozone is absorbed by turpentine oil, therefore, volume of ozone in 100ML of the mixture `=100-90=10mL` `therefore""O_(2)" in the mixture"=100-10=90mL` As 1 L of the mixture WEIGH = 1.5 g, therefore, average MOLAR mass of the mixture (mass of 22.4 L at S.T.P.)`=1.5xx22.4=33.6"g mol"^(-1)` `"Ratio of ozone " : "oxygen in the mixture "=10:90` It m is the molecular mass of ozone, then `"Average mol. mass of mixture "=(10xxm+90xx32)/(100)="33.6 (CALCULATED above)"` `"or"m+288=336 "or"m=48` |
|