A) \[6.02\times {{10}^{23}}\]
B) \[12.04\times {{10}^{22}}\]
C) \[1.806\times {{10}^{23}}\]
D) \[31.80\times {{10}^{28}}\]
Correct Answer: C
Solution :
Molecular mass of \[N{{a}_{2}}C{{O}_{3}}=2\times 23+12+3\times 16=106\] \[\because 106\,g\,N{{a}_{2}}C{{O}_{3}}\]contains \[=3\times 6.023\times {{10}^{23}}\]oxygen atoms \[\therefore \]10.6 g of \[N{{a}_{2}}C{{O}_{3}}\]will contain \[=\frac{3\times 6.023\times {{10}^{23}}}{106}\times 10.6\] \[=18.069\times {{10}^{24}}\] \[=1.806\times {{10}^{23}}\]oxygen atoms
You need to login to perform this action.
You will be redirected in
3 sec
