A) \[16g\,{{O}_{2}}\]
B) \[14\,g\,{{N}_{2}}\]
C) \[2g\,{{H}_{2}}\]
D) \[6\,g\,{{I}_{2}}\]
Correct Answer: C
Solution :
Key Idea: Molecular mass in gram \[=6.023\times {{10}^{23}}\]molecules Molecular mass of oxygen \[=32\text{ }g\] \[\therefore \] 32g of \[{{O}_{2}}=6.023\times {{10}^{23}}\] molecules of \[{{\text{O}}_{\text{2}}}\] \[\therefore \]16 g of \[{{O}_{2}}=0.5\times 6.023\times {{10}^{23}}\] molecules of \[{{O}_{2}}\] (b) Molecular mass of nitrogen = 28 g \[\therefore \]28 g of \[{{N}_{2}}=6.023\times {{10}^{23}}\]molecules of \[{{N}_{2}}\] \[\therefore \]14 g of \[{{N}_{2}}=0.5\text{ }\times \text{ }6.023\times {{10}^{23}}\] molecules of\[{{N}_{2}}\] (c) Molecular mass of hydrogen = 2g \[\therefore \]2 g of \[{{H}_{2}}=6.023\times {{10}^{23}}\]molecules of\[{{H}_{2}}\] Molecular mass of iodine \[=256\text{ }g\] \[\therefore \]256 g of \[{{I}_{2}}=6.023\times {{10}^{23}}\]molecules of \[{{I}_{2}}\] \[\therefore \]\[6\,g\,\]of \[{{I}_{2}}=\frac{6}{256}\times 6.023\times {{10}^{23}}\]of \[{{I}_{2}}\] \[\therefore \]\[2g\]of \[{{H}_{2}}\](choice c) has highest number of molecules.You need to login to perform this action.
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