Question

$25.3$ g of sodium carbonate, $N{a}_{2}C{O}_3$ is dissolved in enough water to make $250$ mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, $N{a}^{+}$ and carbonate ions, $C{O}_3^{2-}$ are respectively.

[Molar mass of $N{a}_{2}C{O}_3=106$g $mo{l}^{-1}$]

• A. $0.477$M and $0.477$M
• B. $0.955$M and $1.910$M
• C. $1.910$M and $0.955$M
• D. $1.90$M and $1.910$M

Answer 1

The correct option is C $1.910$M and $0.955$M

Molar Mass of ${Na}_2{CO}_3$=$106{gmol}^{-1}$

Given Mass=$25.3g$

Volume=$250mL$

No. of moles of ${Na}_2{CO}_3=\frac{GivenMass}{MolarMass}$

$=\frac{25.3}{106}=0.238mole$

$Molarity=\frac{No.ofmoles}{VolumeofSolutioninL}$

$=\frac{0.238}{250\times {10}^{-3}L}=0.954M$

${Na}_2{CO}_3⟶2{Na}^{+}+{CO}_3^{-2}$

$1$mole of ${Na}_2{CO}_3$ gives $2$mole of ${Na}^{+}$ and $1$mole of ${CO}_3^{-2}$

$0.952Mole⟶2\times 0.952moleof{Na}^{+}=1.904M$

$\Rightarrow 1\times 0.952moleof{CO}_3^{-2}=0.952M$

${Na}^{+}=1.904M$

${CO}_3^{-2}=0.952M$

Hence, the correct option is $C$