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Question: Which of the following gives equivalent mass of \(N{a_2}C{O_3}\) when titrated against \(HCl\) in th...

Which of the following gives equivalent mass of Na2CO3N{a_2}C{O_3} when titrated against HClHCl in the presence of methyl orange?
A. 5.3
B. 53
C. 10.6
D. 106

Explanation

Solution

The ratio of Atomic Mass and valency of compound​ is called equivalent mass. To find the equivalent mass first find the change in oxidation number to get the valency of the atom.

Complete step by step answer:
Equivalent mass can be defined as the number of parts by mass of substance that combines with or displaces 1.008 parts by mass of hydrogen or 8.0 parts of oxygen or 35.5 parts of chlorine. In hydrogen displacement method, the equivalent mass is –
mMmH2\dfrac{{{m_M}}}{{{m_{{H_2}}}}}
where, mM{m_M} is the mass of metal and mH2{m_{{H_2}}} is the mass of H2{H_2} displaced.
We know that, when metals react with acids then they can produce hydrogen. To calculate the equivalent mass we have to use the hydrogen displacement. When some mass of an element reacts with an acid then it produces some amount of hydrogen gas under standard temperature and pressure (STP).
Therefore, the reaction is:
Na2CO3+2HClMethyl orange2NaCl+H2O+CO2N{a_2}C{O_3} + 2HCl\mathop \to \limits^{{\text{Methyl orange}}} 2NaCl + {H_2}O + C{O_2}
The valency of compound is 2 because the change in oxidation number in the compound is 2.
Equivalent mass of  Na2CO3=1062=53\therefore Equivalent{\text{ }}mass{\text{ }}of\;N{a_2}C{O_3} = \dfrac{{106}}{2} = 53
Hence, when the equivalent mass of Na2CO3N{a_2}C{O_3} is titrated against HClHCl in the presence of methyl orange then, the equivalent mass is 53.

So, the correct answer is Option B .

Note:
To calculate the equivalent mass of a base, simply divide the molar mass of the base by the number of hydroxyl groups. Take, for example, calcium hydroxide Ca(OH)2Ca{\left( {OH} \right)_2} A few simple calculations yield an equivalent of 37 g/mol. To find the equivalent of an acid, divide the acid's molar mass by the number of protons.