A) 0.955 M and 1.910 M
B) 1.910 M and 0.955 M
C) 1.90 M and 1.910M
D) 0.477 M and 0.477 M
Correct Answer: B
Solution :
[b] Concentration of \[N{{a}_{2}}C{{O}_{3}}=\frac{25.3}{106}\times \frac{1000}{250}=0.955\,M\] \[[N{{a}^{+}}]=2\times 0.955=1.91\,M\] \[\left[ CO_{3}^{2-} \right]=0.955\,M\]You need to login to perform this action.
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