A) 0.955 M and 1.910 M
B) 1.910 M and 0.955 M
C) 1.90 M and 1.910 M
D) 0.477 M and 0.477 M
Correct Answer: B
Solution :
Molarity\[\text{=}\frac{\text{number}\,\text{of}\,\text{moles}\,\text{of}\,\text{solute}}{\text{volume}\,\text{of}\,\text{solution}\,\text{(in}\,\text{mL)}}\text{ }\!\!\times\!\!\text{ 1000}\] \[=\frac{25.3\times 1000}{106\times 250}=0.9547\approx 0.955\,M\] \[N{{a}_{2}}C{{O}_{3}}\]in aqueous solution remains dissociated as \[\underset{x}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,\underset{2x}{\mathop{2N{{a}^{+}}}}\,+\underset{x}{\mathop{CO_{3}^{2-}}}\,\] Since, the molarity of \[N{{a}_{2}}C{{O}_{3}}\]is 0.955 M, the molarity of \[CO_{3}^{2-}\]is also 0.955 M and that of Na+ is 2 x 0.955=1.910 M
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