A) \[5g\,N{{H}_{3}}\]
B) \[11\,g\,C{{O}_{2}}\]
C) \[8\,gS{{O}_{2}}\]
D) \[~4\,g\,{{H}_{2}}\]
Correct Answer: D
Solution :
Key Idea: Molecular mass in gram\[~=1\text{ }mole=6.023\times ~{{10}^{23}}\]molecules, find the moles of all given choices. More the moles, more will be number of molecules in it. (a) 5 g of \[\text{N}{{\text{H}}_{\text{3}}}\]\[(M.wt.\,of\,N{{H}_{3}}=17)\] 17g of \[N{{H}_{3}}=1\,mol\] \[\therefore \] 5 g of \[N{{H}_{3}}=\frac{1}{17}\times 5=0.294\,\text{mol}\] (b) 11 g of \[C{{O}_{2}}\]\[(M.wt.of\,C{{O}_{2}}=44g)\] \[44\,g\,\text{of}\,C{{O}_{2}}=1\,\text{mol}\] \[\therefore \]11 g of \[C{{O}_{2}}=\frac{1}{44}\times 11=0.25\,\text{mol}\] (c)\[8\,g\]of \[S{{O}_{2}}\]\[(M.wt.of\,S{{O}_{2}}=64)\] \[64\,g\]of \[S{{O}_{2}}=1\,\text{mol}\] \[\therefore \]\[8\,g\]of \[S{{O}_{2}}=\frac{1}{64}\times 8=0.125\,\text{mol}\] (d)\[4\,g\]of \[{{H}_{2}}\]\[(M.wt.of\,{{H}_{2}}=2)\] 2 g of \[{{H}_{2}}=1\,\,mol\] \[\therefore \]\[4\,g\]of \[{{H}_{2}}=\frac{1}{24}\times 4=2\,\text{mol}\] \[\therefore \]\[4\,g\]of \[{{H}_{2}}\]has highest moles.You need to login to perform this action.
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