Question

During estimation of nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from $0.5g$ of the compound in Kjeldahl's estimation of nitrogen neutralized $10mL$ of $1M{{H}_{2}}S{{O}_{4}}$​. The percentage of nitrogen in the compound is:
(A) $56%$
(B) $45%$
(C) $50%$
(D) $40%$

Answer 1

Hint:The percentage composition of a substance is the amount of that present in a given molecule, or a material provided total the mass or the volume of the content.
Molarity is a measure to express the concentration of a solution in terms of volume and number of moles.

Complete step-by-step solution:The concentration of all the reagents are expressed in terms of molarity. So, we will consider the definition of molarity which is, number of moles of solute present per litre of the solvent. The molarity can be expressed mathematically as,
$M=\dfrac{n}{V}$
Where $M$ represents the molarity of the solution, $n$ is the number of moles and $V$ is the volume of the solution, in litres.
Now, if we consider the question, we are provided some of the information, let us write the given things first.
$10mL$ of $1M{{H}_{2}}S{{O}_{4}}$ is neutralized by $20mL$ of $1M$$N{{H}_{3}}$
By the definition of molarity, $1L$ of $1M$ ammonia will contain $14g$ of nitrogen, as the molar mass of nitrogen is $14g$. And we can write $1L$ as $1000mL$.
So, by using the unitary method we can calculate the amount of nitrogen contained in $20mL$ of $1M$ ammonia, which is,
$\dfrac{14\times 20}{1000}g$ of nitrogen.
Or, $0.5g$ of the compound contains $\dfrac{14\times 20}{1000\times 0.5}g$ of nitrogen.
In order to calculate the value of nitrogen in terms of percentage, we will calculate the same value for $100g$ of the compound, which is
$\dfrac{14\times 20\times 100}{1000\times 0.5}g$
Now, after solving this equation we get, $56%$ as the nitrogen content.

Hence, the most appropriate answer would be option (A).

Note:The percentage estimation of a specific element in a compound can be done by the help of molarity or the concentration of the compound, as the molarity is the number of moles of solute present per litre of the solvent.
Neutralization is a process in which a base or an acid is added to an acid or a base, in that order, in order to neutralise the hydronium ion concentration in the solution. Here, the sulphuric acid, which acts as an acid was added to ammonia which acts as a base.