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Question: \( 0.98\text{ }g \) of an acid (A) of phosphorus in \( 100\text{ }mL \) solution is \( 0.1\text{ }M ...

0.98 g0.98\text{ }g of an acid (A) of phosphorus in 100 mL100\text{ }mL solution is 0.1 M0.1\text{ }M and neutralizes 300300 mL of 0.1 N NaOH0.1\text{ }N\text{ }NaOH
(A)MgCl2(B)Δ(C)\left( A \right){\xrightarrow{\mathop{MgCl_2}}}\,\left( B \right){\xrightarrow{\mathop{\Delta}}}\,\left( C \right)
If 1.111.11 of (C)\left( C \right) is obtained from given (A).\left( A \right). Compounds (A), (B) and (C) are identified as: A:H3PO4, B:MgHPO4A:{{H}_{3}}P{{O}_{4}},\text{ }B:MgHP{{O}_{4}} ​ and C:Mg2P2O7C:M{{g}_{2}}{{P}_{2}}{{O}_{7}}
​If true enter 11 , else enter 0.0.

Explanation

Solution

Hint : We know that the oxidation number of an atom is the charge that results when the electrons in a covalent bond are assigned to the more electronegative atom. Oxidation state and oxidation number are quantities that commonly equal the same value for atoms in a molecule and are often used interchangeably.

Complete Step By Step Answer:
Let X be the molar mass of acid of phosphorus. The product of the molar mass and molarity is the amount in g/L.g/L.
X molarity= g/L\therefore X\text{ }molarity=\text{ }g/L
Hence, X 0.1= 9.8gL1X\text{ }0.1=\text{ }9.8g{{L}^{-1}}
Molar mass of acid = 98 g mol 1,=\text{ }98\text{ }g\text{ }mo{{l}^{~-1}}, Let basicity of acid be = x,=\text{ }x, by converting the molarity of acid in normality of acid then 0.10.1 M acid =(0.1 X )×N=\left( 0.1\text{ }X\text{ } \right)\times N
The amount of acid present is equal to the amount of the base neutralized.
100× 0.1 XX=300X(0.1 N NaOH)100\times \text{ }0.1\text{ }XX=300X\left( 0.1\text{ }N\text{ }NaOH \right) , Since we know that basicity x=3x=3
Thus, acid is tribasic which is phosphoric acid., H3PO4{{H}_{3}}P{{O}_{4}} .
Hence, the compound A is H3PO4{{H}_{3}}P{{O}_{4}}
The compound B is MgHPO4MgHP{{O}_{4}}
The compound C is Mg2P2O7M{{g}_{2}}{{P}_{2}}{{O}_{7}}
Now we just have to identify if C is confirmed by the following reaction: 2H3PO4Mg2P2O72{{H}_{3}}P{{O}_{4}}\xrightarrow{{}}M{{g}_{2}}{{P}_{2}}{{O}_{7}}
22 moles of A gives One mole of C.
Hence, 0.98 g0.98\text{ }g of A will give 1.11 g1.11\text{ }g of C.
Correct answer is 1.

Additional Information:
Most of the time, it doesn't matter if the term oxidation state or oxidation number is used. Oxidation state refers to the degree of oxidation of an atom in a molecule. Oxidation states are typically represented by integers, which can be positive, negative, or zero. In some cases, the average oxidation state of an element is a fraction, such as 8/38/3 for iron in magnetite

Note :
Remember that Phosphorus belongs to the group 15.15. It has 55 electrons in its valence shell. So it can either lose 5 e 5\text{ }e\text{ }- to attain +5 oxidation state or gain 3 e3\text{ }e- to attain 3-3 oxidation state. Hence its oxidation state varies from 3-3 into +5+5 in H3PO4.{{H}_{3}}P{{O}_{4}}.