25.3 g of sodium carbonate,Na_2CO_3 is disso | vedclass

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(B) Step $1$: Calculate the molarity of $Na_2CO_3$ solution.Molarity $(M) = \frac{	ext{mass of solute}}{	ext{molar mass} 	imes 	ext{volume of solu

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$1.910 \ M$ and $0.955 \ M$

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(B) Step $1$: Calculate the molarity of $Na_2CO_3$ solution.Molarity $(M) = \frac{	ext{mass of solute}}{	ext{molar mass} 	imes 	ext{volume of solution in } L}$$M = \frac{25.3 \ g}{106 \ g \ mol^{-1} 	imes 0.250 \ L} = 0.955 \ M$Step $2$: Write the dissociation equation.$Na_2CO_3(aq) ightarrow 2Na^{+}(aq) + CO_3^{2-}(aq)$Step $3$: Determine the concentration of ions.Since $1 \ mol$ of $Na_2CO_3$ produces $2 \ mol$ of $Na^{+}$ and $1 \ mol$ of $CO_3^{2-}$,$[Na^{+}] = 2 	imes 0.955 \ M = 1.910 \ M$$[CO_3^{2-}] = 1 	imes 0.955 \ M = 0.955 \ M$Thus,the concentrations are $1.910 \ M$ and $0.955 \ M$ respectively.

$25.3 \ g$ of sodium carbonate,$Na_2CO_3$ is dissolved in enough water to make $250 \ mL$ of solution. If sodium carbonate dissociates completely,the molar concentration of sodium ions,$Na^{+}$,and carbonate ions,$CO_3^{2-}$,are respectively:
(Molar mass of $Na_2CO_3 = 106 \ g \ mol^{-1}$)

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$25.3 \ g$ of sodium carbonate,$Na_2CO_3$ is dissolved in enough water to make $250 \ mL$ of solution. If sodium carbonate dissociates completely,the molar concentration of sodium ions,$Na^{+}$,and carbonate ions,$CO_3^{2-}$,are respectively:(Molar mass of $Na_2CO_3 = 106 \ g \ mol^{-1}$)