Question

0.84 g of a metal carbonate reacts exactly with 40 mL of ${N}/{2}\;{{H}_{2}}S{{O}_{4}}$. The equivalent weight of the metal carbonate is-
  (A) 84
  (B) 64
  (C) 42
  (D) 32

Answer 1

Hint: The Equivalent weight is the mass of a given substance that will displace or combine directly or indirectly with 8 parts by mass of oxygen or 1.008 parts by mass of hydrogen or 35.5 parts by mass of chlorine. From the reaction it’s clear that the equivalence of metal carbonate is equal to equivalence of sulfuric acid and from this we could find the equivalent weight of the metal carbonate.

Complete step by step solution:
The given reaction between the metal carbonate and sulfuric acid can be represented as follows
\[{{M}_{x}}{{\left( C{{O}_{3}} \right)}_{y}}+{{H}_{2}}S{{O}_{4}}\to \operatorname{Product}\]
In the question the normality of sulfuric acid is given a 0.5 N. From this the molarity of sulfuric acid can be calculated by multiplying the normality and the valence factor of sulfuric acid. This relation can be represented as follows
\[Normality=Molarity\times Valence\text{ }factor\]
As we know, the valence factor of sulfuric acid is two. Substitute this in the above equation and on rearrangement of the terms we get
\[Molarity\text{ }of\text{ }sulfuric\text{ }acid=\dfrac{0.5}{2}=0.25\]
As we know Molarity can be defined as the moles of a solute per liters of a solution and the equation for molarity in mL can be represented a follows
\[Molarity\text{ }=\dfrac{Moles}{Volume\left( in\text{ }mL \right)}\times 1000\]
 From the above equation we could find the moles of sulfuric acid as follows
\[moles\text{ }of\text{ }sulfuric\text{ }acid moles\text{ }of\text{ }{{H}_{2}}S{{O}_{4}}=\dfrac{0.25\times 4}{1000}=0.01moles\]
The number of moles can also be obtained by dividing the given mass by molar mass. The molar mass of sulfuric acid is 98 gram per mole. The relationship is shown below
\[Moles=\dfrac{Given\text{ }mass}{Molar\text{ }mass}\]
\[Mass\text{ }{{H}_{2}}S{{O}_{4}}=0.01mol\times 98{g}/{mol}\;=0.98g\]
From the reaction it’s clear that the equivalence of metal carbonate is equal to equivalence of sulfuric acid and the equivalence can be calculated as follows.
\[Equivalence=\dfrac{Mass\text{ }of\text{ substance}}{Equivalent\text{ }weight\text{ }of\text{ }that\text{ substance}}\]
The equivalent weight of sulfuric acid is thus$\dfrac{98}{2}$=49 g. We have found the mass of sulfuric acid as 0.98 g. Let the equivalent weight of metal carbonate be x grams. The mass of the metal carbonate is given as 0.84 g. Substitute the above values in the equation we get
\[\dfrac{0.98}{49}=\dfrac{0.81}{x}\]
\[x=42grams\]

Therefore the answer is option (C) 42

Note: The difference between the terms molecular weight and equivalent weight should be noted down. The molecular weight or the relative molecular mass is the mass of a molecule and is calculated as the sum of the atomic weights of each constituent element multiplied by the number of atoms of that element in the molecular formula whereas equivalent weight corresponds to the atomic weight divided by the usual valency