A) \[1\times {{10}^{23}}\]
B) \[2\times {{10}^{23}}\]
C) \[4\times {{10}^{23}}\]
D) \[6\times {{10}^{23}}\]
Correct Answer: D
Solution :
\[\because \] 17gm \[N{{H}_{3}}\]contains \[6\times {{10}^{23}}\]molecules of \[N{{H}_{3}}\] \[\therefore \] 4.25gm \[N{{H}_{3}}\]contains = \[\frac{6\times {{10}^{23}}}{17}\times 4.25\] \[\therefore \] No. of atoms \[=\frac{6\times {{10}^{23}}\times 4.25}{17}\times 4\] \[=6\times {{10}^{23}}\].
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