Question

# One normal solution of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ contains:A.One formula weight per litre.B.Two formula weight per litreC.Half formula weight per litre.D.One third $\left( {\dfrac{1}{3}} \right)$ formula weight per litre.

Verified
130.8k+ views
Hint:
(1)The ‘term normality of a solution’ is used to refer to the number of gram equivalents of the solute dissolved per litre or ${\text{d}}{{\text{m}}^{\text{3}}}$ of the solution. It is represented by the letter N. Mathematically, it is the number of gram equivalent of the given solute divided by the volume of the solution in litres.
${\text{N = }}\dfrac{{{\text{Gramequivalents}}}}{{{\text{Volume(litres)}}}}$
(2) In case of ionic compounds, the formula of the ionic compound does not represent one molecule of the compound. In ionic compounds, one positive ion is surrounded by a number of negatively charged ions and a negative ion is surrounded by a number of positively charged ions. Thus, no true molecule exists for ionic compounds and the formula represents the simple ratio of the positive and the negative ions present in them. Therefore, it is more suitable to use the term ‘formula weight’ for ionic compounds instead of ‘molecular weight’.

We are given one normal of sodium carbonate solution and we need to find out how many formula weights are present in one litre of the sodium carbonate solution.
From the hint, we have the relation between normality and gram equivalent. Now we need to know the relation between gram equivalent and formula weight.
Gram equivalent is the mass in grams which is numerically equal to the equivalent weight. For a salt, equivalent mass is equal to the molecular mass of the salt divided by the total positive valency of the metal atoms.
${\text{Eq}}{\text{.weight}}\left( {{\text{salt}}} \right){\text{ = }}\dfrac{{{\text{Mol}}{\text{.weight}}\left( {{\text{salt}}} \right)}}{{{\text{Total Valency}}}}$
For an ion, it is equal to the formula mass of the ion divided by the charge on the ion.
${\text{Eq}}{\text{.weight}}\left( {{\text{ion}}} \right){\text{ = }}\dfrac{{{\text{Formulaweight}}\left( {{\text{ion}}} \right)}}{{{\text{Charge}}\left( {{\text{ion}}} \right)}}$
Since ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is an ionic compound and total positive charge on the sodium metal ion is 2, its equivalent mass will be:
Equivalent weight ${\text{ = }}\dfrac{{{\text{Formulaweight}}}}{{\text{2}}}$
Now, we have normality equal to equivalent weight by volume in litres. So, for ionic sodium carbonate we will have:
${\text{Normality = }}\dfrac{{{\text{Equivalent Weight}}}}{{{\text{1litre}}}}$
Substitute the equivalent weight expression in the normality expression.
${\text{Normality = }}\dfrac{{\dfrac{{{\text{Formulaweight}}}}{{\text{2}}}}}{{{\text{1litre}}}}$
So, finally we will get:
${\text{Normality = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ }}\dfrac{{{\text{formulaweight}}}}{{{\text{1litre}}}}$
So, one normal solution of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ will contain half formula weight per litre.

Hence, the correct option is C.

Note:
Molality, mole fraction, mass fraction etc. are preferred to molarity and normality because the former involves the masses of the solute and solvent but the latter involves the volume of solutions. Since temperature has no effect on mass but has significant effect on volume, thus, the molality, mole fraction etc. do not change with temperature but molarity and normality change with temperature.