Answer:
\[10\,g\,{{O}_{2}}=\frac{10}{32}\,\text{mole}\]\[=\frac{10}{32}\times
6.02\times {{10}^{23}}\]molecules\[=1\cdot 88\times {{10}^{23}}\] molecules
\[=2\times 1.88\times {{10}^{23}}\] atoms\[=3.76\times
{{10}^{23}}\] atoms
\[10g\,{{O}_{3}}=\frac{10}{48}\]mole\[=\frac{10}{48}\times
6.02\times {{10}^{23}}\] molecules\[=1.254\times {{10}^{23}}\]molecules
\[=3\times
1.254\times {{10}^{23}}\]atoms \[=3.76\times {{10}^{23}}\] atoms
Thus, both contain the same number of
atoms but bulb \[{{B}_{1}}\] contains more number of molecules.
You need to login to perform this action.
You will be redirected in
3 sec