A) \[34g\] of water
B) \[28g\] of \[C{{O}_{2}}\]
C) \[46g\] of \[C{{H}_{3}}OH\]
D) \[54g\] of \[{{N}_{2}}{{O}_{5}}\]
Correct Answer: A
Solution :
A. 34gm of water \[\because \] 18gm \[{{H}_{2}}O\] =\[6.023\times {{10}^{23}}\] molecule \[\therefore \] 34gm \[{{H}_{2}}O\] = \[\frac{6.023\times {{10}^{23}}}{18}\times 34\] \[=11.37\times {{10}^{23}}\]mole B. 28gm of \[C{{O}_{2}}\] \[\because \] 44gm \[C{{O}_{2}}\]\[=6\times {{10}^{23}}\] molecules \ 28gm \[C{{O}_{2}}\] \[=\frac{6\times {{10}^{23}}}{44}\times 28=3.8\times {{10}^{23}}\] C. 46gm of \[C{{H}_{3}}OH\] \[\because \] 32gm \[C{{H}_{3}}OH\ =6\times {{10}^{23}}\] molecules \ 46gm \[C{{H}_{3}}OH\ =\frac{6\times {{10}^{23}}}{32}\times 46=8.625\times {{10}^{23}}\] D. \[\because \] 108gm of \[{{N}_{2}}{{O}_{5}}=6\times {{10}^{23}}\] molecules \ 54gm of \[{{N}_{2}}{{O}_{5}}\ =\frac{6\times {{10}^{23}}}{108}\times 54=3\times {{10}^{23}}\] molecules.
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