Question

When $25.3\,g$ 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 ion $C{O_3}^{2 - }$ are respectively. Given the molar mass of sodium carbonate is $106\,g/mol$.
A)$0.995\,M\,\& \,1.910\,M$
B)$1.910\,M\& 0.995\,M\,$
C)$\,1.90\,M\& 1.910\,M\,$
D)$\,0.477\,M\& 0.477\,M\,$

Answer 1

Hint:We know that the amount of moles in a given amount of any substance is equal to the grams of the substance divided by its molecular weight.
The mathematically expressed as,
$Mole = \dfrac{{weight\,of\,the\,substance}}{{\,Molecular\,weight}}$

Complete step by step answer:

First, calculate the number of moles of sodium carbonate using the formula,
$Mole = \dfrac{{weight\,of\,the\,substance}}{{\,Molecular\,weight}}$
Given the molar mass of sodium carbonate is $106\,g/mol$.
The mass of sodium carbonate is $25.3\,g$.
The volume of the solution is $250\,ml$.
The number of moles of sodium carbonate is,
$Moles = \dfrac{{25.3\,g}}{{106\,g/mol}}$
$Moles = 0.239\,moles$
The number of moles of sodium carbonate is $0.239\,moles$.
We know that,
Molarity is defined as the mass of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,
$Molarity = \dfrac{{Mass\,of\,solute(in\,moles)}}{{Volume\,of\,solution\,(in\,litres)}}$
Now, we calculate the molarity of the solution using the above formula,
$Molarity = \dfrac{{0.239\,mol}}{{0.25\,L}}$
$Molarity = 0.956\,mol/L$
Thus, the molarity of the solution is $0.956\,mol/L$.
Sodium carbonate dissociates into its ions as follows,
$N{a_2}C{O_3}\xrightarrow{{}}2N{a^ + } + C{O_3}^{2 - }$
We know the concentration of carbonate ions is $0.956\,mol/L$.
Since two sodium ions are in solution then the concentration of sodium ion is $2 \times 0.956\,mol/L = 1.912\,mol/L$.
The obtained values are found to be under option B.
Therefore, the option B is correct.

Note:
Don't confuse the terms of molality and molarity. The number of moles of solute in one liter of solution is defined as the molarity and molality is used to measure the moles to the kilogram of the solvent.
The mathematical expression of molality is,
\[Molality{\text{ }}\left( m \right){\text{ }} = {\text{ }}\dfrac{{moles{\text{ }}of{\text{ }}solute\left( {Mol} \right)}}{{kilograms{\text{ }}of{\text{ }}solvent\left( {Kg} \right)}}{\text{ }}\]