1.

The number of atoms of oxygen present in 11.2 L of ozone at N.T.P. are :

Answer»

`3.01 XX 10^23`
`6.02 xx 10^23`
`9.03 xx 10^23`
`1.20 xx 10^24`

Solution :22.4 L of OZONE at N.T.P.` = 6.02 xx 10^23`
molecules of `O_3 = 3 xx 6.02 xx 10^23` ATOMS of O 11.2 L of ozone at N.T.P.
` = (3 xx 6.02 xx 10^23)/(22.4) xx 11.2 = 9.03 xx 10^23`


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