A) \[0.960\]
B) \[96.0\]
C) \[0.0960\]
D) \[9.60\]
Correct Answer: A
Solution :
:\[6.022\times {{10}^{23}}\] molecules are present in 32g of \[{{O}_{2}}\] \[\therefore \] \[1.8\times {{10}^{22}}\]molecules will be present in \[\frac{32}{6.022\times {{10}^{23}}}\times 1.8\times {{10}^{22}}g\] \[=0.960g\] of \[{{O}_{2}}\]
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