A) 34 g of water
B) 28 g of \[C{{O}_{2}}\]
C) 46 g of \[C{{H}_{3}}OH\]
D) 54 g of \[{{N}_{2}}{{O}_{5}}\]
Correct Answer: A
Solution :
34 g of water \[\because \] 18 g of \[{{H}_{2}}O=6\times {{10}^{23}}\] molecule \[\therefore \] 34 g of \[{{H}_{2}}O=\frac{6\times {{10}^{23}}\times 34}{18}\] \[=11.33\times {{10}^{23}}\] molecules 28 g of \[C{{O}_{2}}\] \[\because \] 44 g of \[C{{O}_{2}}=6\times {{10}^{23}}\] molecules \[\therefore \]28 of \[C{{O}_{2}}=\frac{6\times {{10}^{23}}\times 28}{44}\] \[=3.8\times {{10}^{23}}\] molecules 46 g of \[C{{H}_{3}}OH\] \[\because \] 32 g of \[C{{H}_{3}}OH=6\times {{10}^{23}}\] molecules \[\therefore \] 46 g of \[C{{H}_{3}}OH=\frac{6\times {{10}^{23}}\times 46}{32}\] \[\because \] 108 g of \[{{N}_{2}}{{O}_{5}}=6\times {{10}^{23}}\] molecules \[\therefore \] 54 g of \[{{N}_{2}}{{O}_{5}}=\frac{6\times {{10}^{23}}\times 54}{108}\] \[=3\times {{10}^{23}}\] molecules
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