Answer
64.8k+ views
Hint: As we know in a chemical reaction, equivalence of a substance is explained as the amount of which combines with 1 mole of hydrogen atoms or replaces the same number of hydrogen atoms. Thus equivalent weight in grams is weight in grams of 1 equivalent.
Complete step by step answer:
In this question given that ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ disproportionate and form ${\text{P}}{{\text{H}}_{\text{3}}}$ . We can write this equation is as follows:
$
{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}} \to {\text{P}}{{\text{H}}_{\text{3}}} \\
{{\text{P}}^ + } + 4{e^ - } \to {{\text{P}}^{3 - }} \\
$
Thus equivalent weight of $\left( {{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}} \right)$ in ${\text{P}}{{\text{H}}_{\text{3}}}$ is,
$
{\text{ = }}\dfrac{{{\text{Molecular}}\,{\text{weight}}}}{{{\text{Valance}}\,{\text{factor}}}} \\
{\text{ = }}\dfrac{{\text{M}}}{{\text{4}}} \\
$
(Here valance factor is 4)
Again, ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ disproportionate and form ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}$ . We can write this equation is as follows:
$
{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}} \to {{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3} \\
{{\text{P}}^ + } - 2{e^ - } \to {{\text{P}}^{3 + }} \\
$
Thus equivalent weight of \[\left( {{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}} \right)\] in ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}$ is,
$
{\text{ = }}\dfrac{{{\text{Molecular}}\,{\text{weight}}}}{{{\text{Valance}}\,{\text{factor}}}} \\
{\text{ = }}\dfrac{{\text{M}}}{2} \\
$
(Here valance factor is 2)
Now we add equivalence weight of ${\text{P}}{{\text{H}}_{\text{3}}}$ and ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}$ .
$\dfrac{{\text{M}}}{{\text{4}}}{\text{ + }}\dfrac{{\text{M}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{3M}}}}{{\text{4}}}$
Thus option D is correct.
Additional Information: We know that the law of equivalence is one equivalence of an element combined with one equivalent of others. Equivalent weight of an acid in an acid base neutralization reaction is the portion of weight of 1 mole of the acid that can furnish 1 mole of hydrogen ion and equivalent weight of a base is part of weight of one mole of base that can furnish 1 mole of hydroxide ion or accept 1 mole of hydrogen ion.
Note:
As we know a redox reaction that reactant is transformed into product by simultaneous reduction reaction as well as oxidation reaction is termed as disproportionation reaction. For a reaction, say hydrogen peroxide transformed into water and oxygen. In this reaction hydrogen peroxide oxidizes and forms oxygen and it reduces and forms water.
Complete step by step answer:
In this question given that ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ disproportionate and form ${\text{P}}{{\text{H}}_{\text{3}}}$ . We can write this equation is as follows:
$
{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}} \to {\text{P}}{{\text{H}}_{\text{3}}} \\
{{\text{P}}^ + } + 4{e^ - } \to {{\text{P}}^{3 - }} \\
$
Thus equivalent weight of $\left( {{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}} \right)$ in ${\text{P}}{{\text{H}}_{\text{3}}}$ is,
$
{\text{ = }}\dfrac{{{\text{Molecular}}\,{\text{weight}}}}{{{\text{Valance}}\,{\text{factor}}}} \\
{\text{ = }}\dfrac{{\text{M}}}{{\text{4}}} \\
$
(Here valance factor is 4)
Again, ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ disproportionate and form ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}$ . We can write this equation is as follows:
$
{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}} \to {{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3} \\
{{\text{P}}^ + } - 2{e^ - } \to {{\text{P}}^{3 + }} \\
$
Thus equivalent weight of \[\left( {{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}} \right)\] in ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}$ is,
$
{\text{ = }}\dfrac{{{\text{Molecular}}\,{\text{weight}}}}{{{\text{Valance}}\,{\text{factor}}}} \\
{\text{ = }}\dfrac{{\text{M}}}{2} \\
$
(Here valance factor is 2)
Now we add equivalence weight of ${\text{P}}{{\text{H}}_{\text{3}}}$ and ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_3}$ .
$\dfrac{{\text{M}}}{{\text{4}}}{\text{ + }}\dfrac{{\text{M}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{3M}}}}{{\text{4}}}$
Thus option D is correct.
Additional Information: We know that the law of equivalence is one equivalence of an element combined with one equivalent of others. Equivalent weight of an acid in an acid base neutralization reaction is the portion of weight of 1 mole of the acid that can furnish 1 mole of hydrogen ion and equivalent weight of a base is part of weight of one mole of base that can furnish 1 mole of hydroxide ion or accept 1 mole of hydrogen ion.
Note:
As we know a redox reaction that reactant is transformed into product by simultaneous reduction reaction as well as oxidation reaction is termed as disproportionation reaction. For a reaction, say hydrogen peroxide transformed into water and oxygen. In this reaction hydrogen peroxide oxidizes and forms oxygen and it reduces and forms water.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)