Question

The molecular mass of organic acid was determined by the study of its barium salt $4.290\,g$ of salt was quantitatively converted to the free acid by the reaction with $21.64\,ml$ of $0.477\,M\,{H_2}S{O_4}$. The barium salt was found to have two moles of water of hydration per $B{a^{2 + }}$ ion and acid is monobasic. What is the molecular weight of anhydrous acid?
A. $130$
B. $142$
C. $100$
D. $110$

Answer 1

Hint: We know that the amount of moles in a given amount of any substance is equal to the grams of the substance divided by its molecular weight.
The mathematically expressed as,
$Mole = \dfrac{{weight\,of\,the\,substance}}{{\,Molecular\,weight}}$
Formula used: We can calculate the molecular weight of substance by using the below mentioned formula.
$Molecular\,weight = \dfrac{{weight\,of\,the\,substance}}{{\,Moles}}$

Complete step by step answer: Given data
The weight of barium salt is $4.290\,g$.
The volume of acid is $21.64\,ml$.
The molarity of the solution is $0.477\,M$.
We know that,
Molarity is defined as the mass of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. We can calculate the molarity of the substance by using the below mentioned formula.
$Molarity = \dfrac{{Mass\,of\,solute(in\,moles)}}{{Volume\,of\,solution\,(in\,litres)}}$
The number of moles can be determine by using the above formula,
$0.477\,M = \dfrac{{Mass\,of\,solute(in\,moles)}}{{{\text{0}}{\text{.02164}}\,L}}$
$Moles\, = 0.477\,M \times {\text{0}}{\text{.02164}}\,L$
Multiplying the above values we get,
$Moles\, = 0.010\,moles$
The molecular weight can be calculated by rearranging the mole formula,
$Molecular\,weight = \dfrac{{weight\,of\,the\,substance}}{{\,Moles}}$
$Molecular\,weight = \dfrac{{4.290g}}{{0.01\,mol}} = 429\,g/mol$
The molecular weight of salt is$429\,g/mol$.
We can write the balanced equation for the given reaction is,
${\left( R \right)_2}B{a^{2 + }}.2{H_2}O + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + 2\left( R \right) + 2{H_2}O$
We know, the molar mass of barium is $137.3\,g/mol$
We must remember that the molecular weight of water is $18\,g/mol$.
Now we can calculate the molecular weight of anhydrous acid by substituting the molecular weight of water and molar mass of barium.
$
  Molecular\,Weight = 429 - \left( {137.3 - 2 \times 18 + 2} \right) \times \dfrac{1}{2} \\
    \\
 $
$Molecular\,Weight = 129\,g$
The molecular weight of anhydrous acid is $129\,g$ which is closely related to option A.
So, the correct answer is “Option A”.

Note: Don't confuse the terms of molality and molarity. The number of moles of solute in one liter of solution is defined as the molarity and molality is used to measure the moles to the kilogram of the solvent
The mathematical expression of molality is,
\[Molality{\text{ }}\left( m \right){\text{ }} = {\text{ }}\dfrac{{moles{\text{ }}of{\text{ }}solute\left( {Mol} \right)}}{{kilograms{\text{ }}of{\text{ }}solvent\left( {Kg} \right)}}{\text{ }}\]