Question

In Duma’s method for estimation of nitrogen, 0.25 g of an organic compound gave 40 ml of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:
(A)- 16.76
(B)- 15.76
(C)- 17.36
(D)- 18.20

Answer 1

Combining Boyle’s law, Charle’s law, Gay-Lussac’s law, and Avagadro’s law gives us the Combined Gas law which can combine into one proportion as-
$V\propto \dfrac{T}{P}$
Removing the proportionality and inserting a constant,
$\dfrac{PV}{T}=C$
This clearly says that as the pressure rises, the temperature also rises and vice versa.
Therefore, we can also say that, $\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}=\dfrac{{{P}_{3}}{{V}_{3}}}{{{T}_{3}}}...$

Complete answer:
-Robert Boyle in 1662 formulated a law known as Boyle’s law or Pressure-Volume law which states that ‘the volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are kept constant. That is,
$V\propto \dfrac{1}{P}$
Removing the proportionality by inserting a constant-
$PV=C$
This clearly says that when the pressure rises, the volume goes down and when volume rises, the pressure goes down.
Therefore, we can also say that, ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}={{P}_{3}}{{V}_{3}}...$
-Jacques Charles in the year 1787 formulated a law known as Charle’s law which states that ‘the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature. That is,
$V\propto T$
Removing the proportionality by inserting a constant,
$\dfrac{V}{T}=C$
This clearly says that as the volume goes up, the temperature also goes up and vice versa.
Therefore, we can also say that, $\dfrac{{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{V}_{2}}}{{{T}_{2}}}=\dfrac{{{V}_{3}}}{{{T}_{3}}}...$
-Joseph Gay-Lussac in the year 1808 gave a law known as Gay-Lussac’s law which states that ‘the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. That is,
$P\propto T$
Removing the proportionality by inserting a constant,
$\dfrac{P}{T}=C$
This clearly says that an increase in temperature pressure will also go up and vice versa.
Therefore, we can also say that, $\dfrac{{{P}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}}{{{T}_{2}}}=\dfrac{{{P}_{3}}}{{{T}_{3}}}...$
-Amedeo Avagadro in 1811 gave a law known as Avagadro’s law, which gives the relationship between volume and amount when pressure and temperature are held constant. This law states that ‘if the amount of gas in a container is increased, the volume increases and vice versa. That is,
$V\propto n$
Removing the proportionality and inserting a constant,
$\dfrac{V}{n}=C$
This means that the volume-amount dfraction will always be the same value if the pressure and temperature remain constant.
Therefore, we can also say that, $\dfrac{{{V}_{1}}}{{{n}_{1}}}=\dfrac{{{V}_{2}}}{{{n}_{2}}}=\dfrac{{{V}_{3}}}{{{n}_{3}}}...$
-According to the question-
$\begin{aligned} & m=0.25g \\\ & {{V}_{1}}=40mL \\\ & {{V}_{2}}=? \\\ & {{T}_{1}}=300K \\\ & {{T}_{2}}=273K \\\ & {{P}_{1}}=725-25=700mm \\\ & {{P}_{2}}=760mm \\\ \end{aligned}$
Inserting the values in the formula of Combined Gas Law,
$\dfrac{700\times 40}{300}=\dfrac{760\times {{V}_{2}}}{273K}$
Finding for the value of ${{V}_{2}}$by rearranging the equation, we get
${{V}_{2}}=\dfrac{{{P}_{1}}{{V}_{1}}{{T}_{2}}}{{{T}_{1}}{{P}_{2}}}=\dfrac{700\times 40\times 273}{300\times 760}=33.52mL$
-Now finding the mass of nitrogen which corresponds to 33.53 ml-
22400 mL of ${{N}_{2}}$at STP weighs = 28 g
Therefore, 33.52 mL of ${{N}_{2}}$ at STP will weigh $=\dfrac{28\times 33.52}{22400}=0.0419g$
-Calculating the mass percent of Nitrogen,
$$
$\Rightarrow$

So, the correct answer is Option A.

Note:
The Dumas method is an analytical method which is used for the qualitative determination of nitrogen in chemical substances. Dumas method was described in the year 1826 by Jean-Baptiste Dumas which is based on the fact that nitrogenous compounds when heated with cupric oxide in an atmosphere of carbon dioxide yields free nitrogen. The general chemical reaction for the Dumas method is given as-
$\begin{aligned} & C+2CuO\to C{{O}_{2}}+2Cu \\\ & 2H+CuO\to {{H}_{2}}O+Cu \\\ & \\\ \end{aligned}$
In some cases, the traces of nitrogen oxide may be formed which are reduced to elemental nitrogen by passing the overheated copper spiral.
$\begin{aligned} & Nitrogen+CuO\to {{N}_{2}}+\text{Oxides of nitrogen} \\\ & \text{Oxides of nitrogen}+Cu\to {{N}_{2}}+CuO \\\ \end{aligned}$
The mixture of gases was collected over a KOH solution which absorbs carbon dioxide.