Question

What is the molarity and normality of a $13$% solution (by weight) of sulphuric acid with a density of $1.02$ g/mL and also calculate the volume (in mL) that is required to prepare $1.5$N of this acid to make it up to $100$mL?
A. Molarity =$1.35$, Normality =$2.70$, Volume = $180$mL
B. Molarity =$2$, Normality =4, Volume = 200mL
C. Molarity =, Normality =$2$ $3$, Volume = $150$mL
D. None of these

Answer 1

First we will determine the mole of sulphuric acid then by using density formula we will determine the volume of solution. We use a molarity formula to determine the molarity. We will determine the f-factor for sulphuric acid and then multiplying the n-factor with molarity we will determine the normality. By using the dilution relation of normality and volume we will determine the volume.

Formula used:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{\,{\text{molar}}\,{\text{mass}}}}$
${\text{density}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{volume}}}}$
${\text{molarity}}\,\,{\text{ = }}\,\dfrac{{{\text{moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{volume}}\,{\text{of}}\,{\text{solution}}}}$
Normality = ${\text{molarity}}\,\,{\text{ \times }}\,{\text{valence factor}}$
${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}\,{\text{ = }}\,{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}$

Complete step-by-step answer:
$13$% (by weight) means $13$g sulphuric acid is dissolved in $100$g of solution.
So, the amount of solvent is,
$100 - 13\, = \,87$g.
We will determine the mole of sulphuric acid as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{\,{\text{molar}}\,{\text{mass}}}}$
Molar mass of sulphuric acid is $98$ g/mol.
On substituting $13$g for mass and $98$ g/mol for molar mass,
${\text{mole}}\,{\text{ = }}\,\dfrac{{13\,{\text{g}}}}{{\,98\,{\text{g/mol}}}}$
${\text{mole}}\,{\text{ = }}\,0.133\,{\text{mol}}$
So, the mole of sulphuric acid is$0.133$mole.
We will convert the amount of solvent from g to kg as follows:
${\text{1000}}\,{\text{g = }}\,{\text{1}}\,{\text{kg}}$
$87\,{\text{g}}\,{\text{ = }}\,0.087\,{\text{kg}}$

We will determine the volume of water as follows:
${\text{density}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{volume}}}}$
Substitute $1.02\,{\text{g}}\,{\text{. m}}{{\text{L}}^{ - 1}}$ for density of solution and ${\text{100g}}$ for mass of solution.
$1.02\,{\text{g}}\,{\text{. m}}{{\text{L}}^{ - 1}}\,{\text{ = }}\,\dfrac{{100\,{\text{g}}}}{{{\text{volume}}}}$
${\text{volume}}\,{\text{ = }}\,\dfrac{{100\,{\text{g}}}}{{1.02\,{\text{g}}\,{\text{. m}}{{\text{L}}^{ - 1}}}}$
${\text{volume}}\,{\text{ = }}\,98.04\,{\text{mL}}$
We will convert the volume of water from mL to L as follows:
$1000\,{\text{mL}}\,{\text{ = }}\,{\text{1 L}}$
$98.04\,{\text{mL}} = 0.09804\,{\text{L}}$
We will determine the molarity as follows:
${\text{molarity}}\,\,{\text{ = }}\,\dfrac{{{\text{moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{volume}}\,{\text{of}}\,{\text{solution}}}}$
On substituting $0.09804$L for volume of solution and $0.133$mol for mole of sulphuric acid.
${\text{Molarity}}\,{\text{ = }}\,\dfrac{{0.133\,{\text{mol}}}}{{0.09804\,\,{\text{L}}}}$
${\text{Molarity}}\,{\text{ = }}\,1.35\,{\text{M}}$
So, the molarity is$1.35$M.
Sulphuric acid dissociates as follows:
${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, \to \,2{{\text{H}}^ + }\, + \,{\text{SO}}_4^{2 - }$
Sulphuric acid is given two protons, so the n-factor for the sulphuric acid is $2$.
We will determine the normality of sulphuric acid solution as follows:
${\text{Normality = }}\,{\text{molarity}}\,{\text{ \times }}\,{\text{n}} - {\text{factor}}$
On substituting $2$ for n-factor and $1.35$M for molarity,
${\text{Normality = }}\,{\text{1}}{\text{.35}}\,{\text{ \times }}\,2$
${\text{Normality = }}\,2.70$
So, the normality of the solution is $2.70$N.
Now we will determine the volume of$1.5$N of acid to make it up to $100$mL having normality$2.70$N.
${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}\,{\text{ = }}\,{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}$
Where,
${{\text{N}}_{\text{1}}}$is the normality of the solution having ${{\text{V}}_{\text{1}}}$volume.
${{\text{N}}_{\text{2}}}$is the normality of the solution having ${{\text{V}}_{\text{2}}}$volume.
On substituting $1.5$N for ${{\text{N}}_{\text{1}}}$, $2.70$ N for ${{\text{N}}_{\text{2}}}$, and $100$mL for ${{\text{V}}_{\text{2}}}$,

${\text{1}}{\text{.5}}\,{\text{N}}\,\, \times \,{{\text{V}}_1}\,{\text{ = }}\,{\text{2}}{\text{.70}}\,{\text{N}}\,\, \times \,\,\,{\text{100 mL}}$
${{\text{V}}_1}\,{\text{ = }}\,\dfrac{{{\text{2}}{\text{.70}}\, \times \,{\text{100}}}}{{1.5}}$
${{\text{V}}_1}\,{\text{ = }}\,180\,{\text{mL}}$
So, the volume (in mL) that is required to prepare $1.5$N of this acid to make it up to $100$mL is $180$mL.

Therefore, option (A) Molarity =$1.35$, Normality =$2.70$, Volume = $180$mL, is correct.

Note: n-factor is the number of electrons gained or losses during a reaction for the proton donated. For acids the n-factor is determined as the number of protons donated. Molarity is defined as the moles of solute present in per litter so the solution. Normality is the equivalent weight of solute present in per litter of the solvent. The equivalent weight is determined by dividing the molar mass by n-factor. The solution whose concentration is determined in normality terms is known as normal solution where the solution whose concentration is determined in terms of molarity is known as molar solution.