Question

Molecular mass of an organic compound was determined by the study of its Barium salt. The Barium salt contains two moles of water of hydration per $B{{a}^{2+}}$ ion. 2.562 g of this salt reacts completely with 30 ml of $0.2M$ ${{H}_{2}}S{{O}_{4}}$ to produce Barium Sulfate and acid. The acid is monobasic. What is a molecular mass of anhydrous acid? (Gram atomic mass of Ba = 137g)
a.) 128 g/mol
b.) 132 g/mol
c.) 156 g /mol
d.) 168 g/mol

Answer 1

Hint: This question is based on the stoichiometric calculation of chemistry. Compounds always react in the molar ratio of their stoichiometric coefficients. That’s why if we have the information about one substance present in a chemical reaction we can do the calculation for the other substances based on stoichiometric coefficients.

Complete Solution :
Since acid is monobasic that’s why its molecular formula will be HA type where anion will be univalent. Therefore molecular formula of Barium salt will be $Ba{{A}_{2}}$
- Reaction between acid and salt:
$Ba{{A}_{2}}(aq)+{{H}_{2}}S{{O}_{4}}(aq)\to BaS{{O}_{4}}+2{{H}_{2}}O$
According to reaction stoichiometry
Number of moles of Barium salt = Number of moles of sulfuric acid

Millimoles of ${{H}_{2}}S{{O}_{4}}$ = $30\times 0.2$
Millimoles of ${{H}_{2}}S{{O}_{4}}$ = $6$
As, 6 millimoles of sulfuric acid are completely reacting with Barium salt therefore,

Number of moles of Barium salt = 6 millimoles
Given weight of hydrous salt of Barium = 2.562 g
Gram molecular weight of hydrous salt of Barium = weight / moles
Gram molecular weight of hydrous salt of Barium = $\dfrac{2.562}{6\times {{10}^{-3}}}$
Gram molecular weight of hydrous salt of Barium = 427g

As each molecule of Barium salt is containing 2 water molecules therefore,
Gram molecular weight of anhydrous salt of Barium= 427 - 2(18)
Gram molecular weight of anhydrous salt of Barium= 427 - 36
Gram molecular weight of anhydrous salt of Barium= 391 g

Gram molecular weight of anhydrous salt of Barium= ${{M}_{B{{a}^{2+}}}}+2{{M}_{{{A}^{-}}}}$
${{M}_{B{{a}^{2+}}}}+2{{M}_{{{A}^{-}}}}=391$
$137+2{{M}_{{{A}^{-}}}}=391$
$2{{M}_{{{A}^{-}}}}=391-137$
$2{{M}_{{{A}^{-}}}}=254$
${{M}_{{{A}^{-}}}}=127g$
Gram ionic weight of each anion is 127 g and since it is univalent so it will have only one hydrogen ion with it.
Gram molecular weight of organic acid = mass of anion + mass of hydrogen ion
Gram molecular weight of organic acid = 127 + 1
Gram molecular weight of organic acid = 128 g
So, the correct answer is “Option A”.

Additional Information:
${{H}_{2}}S{{O}_{4}}$ is a very strong dibasic acid as well as oxidizing reagent. Its gram molecular weight is 98 and it is a very important chemical present in the laboratory used frequently in qualitative analysis.

Note: Monobasic acids are the acids that give only one ${{H}^{+}}$ when put into the water.
During the crystallization of a compound some molecules of water bind themselves with the compound.
- molecules that remain in the crystal structure are known as water of hydration or water of crystallization. By heating a compound this water can be expelled from the crystal and it becomes anhydrous.