Question

A sample of crystalline $Ba{\left( {OH} \right)_2}.x{H_2}O$ weighing $1.578g$ was dissolved in water and the solution required $40ml$ of 0.25{\text{ }}N$$$$HN{O_3} for complete reaction. The number of molecules of water of crystallization in base are
A) $7$
B) $8$
C) $6$
D)$5$

Answer 1

As we know that in chemistry, the concentration of the solution is very important. The strength of the solution is very important for inhale solution and medicinal chemistry. There are different ways of representing the concentration of solutions. There are molality, molarity, normality formality, mole fraction, mass percentage, volume percentage, mass by volume and parts per million.
Formula used:
The normality of the solution depends on the number of the gram equivalents of solute and the volume of the solution in litres. The normality of the solution is equal to the ratio of the number of the gram equivalent of the solute to the volume of the solution in litres. The symbol of normality is N.
${\text{Normality = }}\dfrac{{{\text{number of gram equivalent of the solute}}}}{{{\text{volume of the solution litre}}}}$
Moles are defined as the given mass of the molecule is divided by the molecular mass of the molecule.
${\text{moles}}\,{\text{ = }}\dfrac{{{\text{mass}}{\text{of molecule}}}}{{{\text{molecular weight of the molecule}}}}$
The molecular weight of the molecule is dependent on the atomic weight of the atom present in the molecule. The molecular weight of the molecule is equal to the sum of the molecular weight of the atom and the number of the respective atom in the molecule.
${\text{Molecular Weight = Number Theatoms \times Atomic Weight Of The Atom}}$

Complete answer:
The given data is
A sample of crystalline $Ba{\left( {OH} \right)_2}.x{H_2}O$ weighing $1.578g$ was dissolved in water and the solution required $40ml$ of 0.25N$$$$HN{O_3} for complete reaction.
The chemical formula of barium hydroxide is $Ba{\left( {OH} \right)_2}.x{H_2}O$.
The chemical reaction for the above discussion is given below,
$Ba{\left( {OH} \right)_2} + 2HN{O_3} \to Ba{(N{O_3})_2} + 2{H_2}O$
The weight of crystalline barium hydroxide is $1.578g$.
The volume of nitric acid consumed in the reaction is $40ml$.
The normality of nitric acid consumed in the reaction is $0.25N$.
By applying law of equivalent is,
Number of equivalent of $Ba{\left( {OH} \right)_2}$= Number of equivalent of $HN{O_3}$
Number of equivalent of Ba{\left( {OH} \right)_2}$$$$ = 0.04 \times 0.25 = 0.01
${V_1}{N_1} = {V_2}{N_2}$
${\text{Normality = }}\dfrac{{{\text{number of gram equivalent of the solute}}}}{{{\text{volume of the solution litre}}}} \times {\text{volume of the solution litre}}$
The gram equivalent of Ba{\left( {OH} \right)_2}$$$$ = 0.01
The number of moles of $Ba{\left( {OH} \right)_2}$=$= \dfrac{{0.01}}{2} = 0.005$
${\text{Moles}}\,{\text{ = }}\dfrac{{{\text{mass}}\,{\text{ofthe}}{\text{molecule}}}}{{{\text{molecular weight of the molecule}}}}$
We can change the formula for our concern
${\text{mass}}\,{\text{of}}\,{\text{the}}\,{\text{molecule}} = {\text{moles}}\, \times {\text{molecular}}\,{\text{weight}}\,{\text{of}}\,{\text{the}}\,{\text{molecu}}le$
The molecular weight of $Ba{\left( {OH} \right)_2}$ is $176.3$.
$= 0.005 \times 176.3 = 0.8565$
The mass of $Ba{\left( {OH} \right)_2}$ is $0.8565$g.
The weight of crystalline $Ba{\left( {OH} \right)_2}.x{H_2}O$ is $1.578g$.
The mass of the water is calculated as,
$= 1.578{\text{ - }}0.8565 = 0.7215$
The mass of the water $Ba{\left( {OH} \right)_2}.x{H_2}O$ is $0.7215$g.
The moles of water is calculated as,
The molecular weight of water is $18$.
${\text{Moles}}\,{\text{ = }}\dfrac{{{\text{mass}}\,{\text{ofthe}}{\text{molecule}}}}{{{\text{molecular weight of the molecule}}}}$
Now we can substitute the known values we get,
$= \dfrac{{0.7215}}{{18}} = 0.04moles$
The number of molecules of water of crystallization in base is calculated as,
$= \dfrac{{0.04}}{{0.005}} = 8$
According to the above discussion, we conclude a sample of crystalline $Ba{\left( {OH} \right)_2}.x{H_2}O$ weighing $1.578g$ was dissolved in water and the solution required $40ml$of 0.25N$$$$HN{O_3} for complete reaction. The number of molecules of water of crystallization in base is $8$.

Hence, option B is the correct answer.

Note:
We have to know that the gram equivalent of the acid is dependent on the molecular mass of the acid divided by the basicity of the acid. The basicity of the acid is nothing but the number of hydrogen ions able to donate the acid in aqueous medium. The gram equivalent of the acid is equal to the ratio of the molecular mass of the acid to the basicity of the acid. In the acid base concept the hydrogen ion is very important. The acids are proton or hydrogen ion donors. The bases are hydrogen ions or proton acceptor in the reaction. The basicity of the acid and the acidity of the base is dependent on the hydrogen ion in the molecule. The molecular weight of the molecule is the sum of the atomic weight of the atoms in the molecule.