A) \[1.5057\times {{10}^{24}}\]
B) \[2.0478\times {{10}^{24}}\]
C) \[3.0115\times {{10}^{24}}\]
D) \[4.0956\times {{10}^{24}}\]
Correct Answer: C
Solution :
\[\because \] 100gm \[CaC{{O}_{3}}\ =6.023\times {{10}^{23}}\]molecules \[\therefore \] 10gm \[CaC{{O}_{3}}\]=\[\frac{6.023\times {{10}^{23}}}{100}\times 10\] \[=6.023\times {{10}^{22}}\] molecule 1 molecule of \[CaC{{O}_{3}}\]= 50 protons \[6.023\times {{10}^{22}}\]molecule of \[CaC{{O}_{3}}\]\[=50\times 6.023\times {{10}^{22}}\] \[=3.0115\times {{10}^{24}}\]
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