Question

Given,${\text{28ml}}$of ${\text{0}}{\text{.1m}}$ oxalic acid solution requires ${\text{10ml}}$ of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ for titration. ${\text{10ml}}$ of this sample of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ when added to an excess of ${\text{N}}{{\text{H}}_{\text{2}}}{\text{OH}}$ liberates ${{\text{N}}_{\text{2}}}$ at STP. Volume of ${{\text{N}}_{\text{2}}}$ liberated at STP?

Answer 1

We need to know that the air is nothing but a mixture of the gases. The mixture of the gas in the air are major in the form of nitrogen, oxygen, carbon dioxide and other gases. In the atmosphere approximately $78\%$ of nitrogen, $21\%$ of oxygen and remaining $1\%$ of other gases in the world. The symbol of nitrogen is ${\text{N}}$ and oxygen is ${\text{O}}$. The natural form of nitrogen and oxygen gases are diatomic. The symbol of diatomic nature of nitrogen and oxygen gases are ${{\text{N}}_2}$ and ${{\text{O}}_{\text{2}}}$.
Formula used:
The ideal gas equation depends on the pressure, temperature, number of moles, volume of the gas molecules in ideal condition.
The ideal gas equation is,
${\text{PV = nRT}}$
Here, the pressure of the gas is P
The volume of the gas is V
The temperature of the gas in kelvin is T
Gas constant is R
The number of moles of the Gas molecules is n
Formula for convert degree Celsius to kelvin in temperature
${\text{kelvin = degree + 273}}$
The product of molality of the acid and volume is equal to the product of molality of the base and volume.
The molality of the solution depends on the moles of the solute and weight of the solution.

Complete answer:
The given data is
$28ml$ of $0.1m$ oxalic acid solution requires $10ml$ of $KMn{O_4}$ for titration.
The product of molality of the acid and volume is equal to the product of molality of the base and volume.
According to the above discussion, we calculate the molality of $N{H_2}OH$.
$28 \times 0.1 = 10 \times m$
$m = 0.28m$
We calculated the molality of $N{H_2}OH$ is $0.28m$.
The molality of the solution depends on the moles of the solute and weight of the solution.
According to the above discussion, we calculate the moles of $N{H_2}OH$.
$Molality = \dfrac{{{\text{Moles of the molecule}}}}{{{\text{Molecular weight of solvent}}}}$
$0.28\, = \dfrac{{{\text{Moles of the molecule}}}}{{0.01}}$
${\text{Moles of the molecule}} = 0.01 \times 0.28$
$= 0.0028moles$
We calculated the moles of $N{H_2}OH$ is $0.0028moles$
$10ml$ of this sample of $KMn{O_4}$ when added to an excess of $N{H_2}OH$ liberates ${N_2}$ at STP.
The ideal gas equation depends on the pressure, temperature, number of moles, and volume of the gas molecules in ideal condition.
The ideal gas equation is,
$PV = nRT$
According to the above discussion, we calculate the volume of ${N_2}$ at STP.
Temperature at degree = $25^\circ C$
Temperature at kelvin = $25 + 273 = 298K$
Gas constant, R= $0.08206Latm/kmol$
The pressure of the gas is$1$ atm
The number of moles is $0.0028moles$
${\text{PV = nRT}}$
We change the formula for calculate the volume of the gas,
$V = \dfrac{{nRT}}{P}$
We substitute the known values in formula
$V = \dfrac{{0.0028 \times 0.08206 \times 298}}{1}$
$V = 68L$
We calculate the volume of ${N_2}$ at STP is $68ml$.
According to the above calculation, we conclude $28ml$ of $0.1m$ oxalic acid solution requires $10ml$of $KMn{O_4}$ for titration. $10ml$ of this sample of $KMn{O_4}$ when added to an excess of $N{H_2}OH$liberates ${N_2}$ at STP. Volume of ${N_2}$ liberated at STP is $68ml$.

Note:
We need to know that air is the one of the most important things in our life. In the world, without air nothing is surveyed. For human beings, breathing purpose air is the important thing for inhaling and exhaling. For human respiration, oxygen is used for inhalation of breathing and carbon dioxide is exhaled gas. This inhale and exhale gases are important for human lifestyle to survey in the world. Each and every gas has unique properties and unique nature in the atmosphere. One gas is not able to replace another gas in the environment.