Question

The Henry's law constant for the solubility of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ a gas in water at $\text{ 298K }$is$\text{ 1}\text{.0}\times \text{1}{{\text{0}}^{\text{5}}}\text{ atm }$. The mole fraction of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ in air is$\text{ 0}\text{.8 }$. The number of moles of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ from air dissolved in 10 moles of water at $\text{ 298K }$ and $\text{5 atm}$ pressure is:
A) $\text{ 4}\text{.0 }\times \text{1}{{\text{0}}^{-4}}\text{ }$
B) $\text{ 4}\text{.0 }\times \text{1}{{\text{0}}^{-5}}\text{ }$
C) $\text{ 5}\text{.0 }\times \text{1}{{\text{0}}^{-4}}\text{ }$
D) $\text{ 4}\text{.0 }\times \text{1}{{\text{0}}^{-6}}\text{ }$

Answer 1

Hint: The pressure has a remarkable effect on the solubility of the gas. The relation between the partial pressure and mole fraction of a gas solute is given as,
$\text{ }{{\text{p}}_{\text{2}}}\text{ = k}{{\chi }_{2}}\text{ }$
Where \[{{\chi }_{2}}\]the mole fraction of solute, and subscript 2 is used for the gas solute. The k is called Henry’s law constant. For a mixture of gases, henry’s law is applicable for each gas independent of the presence of the other gases.

Complete answer:
The solubility of gases at a given temperature increases directly as the pressure. This conclusion forms a basis of what is known as Henry’s law, which may be stated as below,
The mass of a gas dissolved per unit volume of a solvent is proportional to the pressure of the gas in equilibrium with the solution at a constant temperature.
The henry law can be alternatively stated as the pressure of the gas over the solution in which the gas is dissolved is proportional to the mole fraction of the gas dissolved in solution. That is,
$\text{ }{{\text{p}}_{\text{2}}}\text{ = k}{{\chi }_{2}}\text{ }$ ………………………..(1)
Where $\text{ }{{\chi }_{2}}\text{ }$ the mole fraction of solute and subscript 2 is used for the gas solute.
We have given the following data:
The henry law constant k is, $\text{ 1}\text{.0}\times \text{1}{{\text{0}}^{\text{5}}}\text{ atm }$
Mole fraction of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ in water is, $\text{ }{{\chi }_{{{\text{N}}_{\text{2}}}}}=\text{ 0}\text{.8 }$
The number of moles of water, $\text{ }{{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O }}}\text{= 10 mole }$
The pressure of a gas is, $\text{ P = 5 atm }$
To find, the number of moles of nitrogen $\text{ }{{\text{n}}_{{{\text{N}}_{\text{2}}}}}\text{ }$
The partial pressure of the nitrogen gas is,
$\text{ }{{\text{p}}_{\text{2}}}\text{ = }{{\text{P}}_{\text{Total}}}\text{ }{{\chi }_{\text{2}}}\text{ = 5 atm }\times \text{ 0}\text{.8 = 4 atm }$
Now, substitute the value of partial pressure in (1), henry law, we have,
$\begin{align}
  & \text{ }{{\text{p}}_{\text{2}}}\text{ = k}{{\chi }_{2}}\text{ } \\
 & \Rightarrow {{\chi }_{2}}\text{ = }\dfrac{{{\text{p}}_{\text{2}}}}{\text{k}}\text{ = }\dfrac{4\text{ atm}}{1\times {{10}^{5}}\text{atm}}\text{ = }4\text{ }\times {{10}^{-5}}\text{ } \\
\end{align}$
Now, we know that the mole fraction for the gas solute is written as follows,
$\text{ }{{\chi }_{2}}\text{ = }\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}\text{ }\approx \text{ }\dfrac{{{n}_{2}}}{{{n}_{1}}}\text{ }$
Because we are assuming that the number of moles of gas like $\text{ }{{\text{N}}_{\text{2}}}\text{ }$is negligible in comparison with the moles of the solvent.
Let's find out the number of moles of nitrogen$\text{ }{{\text{N}}_{\text{2}}}\text{ }$.
$\begin{align}
  & \text{ }{{\chi }_{{{\text{N}}_{\text{2}}}}}\text{ = }\dfrac{{{n}_{{{\text{N}}_{\text{2}}}}}}{{{n}_{{{\text{H}}_{\text{2}}}\text{O}}}}\text{ } \\
 & \Rightarrow {{n}_{{{\text{N}}_{\text{2}}}}}={{\chi }_{{{\text{N}}_{\text{2}}}}}\text{ }\times \text{ }{{n}_{{{\text{H}}_{\text{2}}}\text{O}}} \\
 & \Rightarrow {{n}_{{{\text{N}}_{\text{2}}}}}=\text{ 4 }\times \text{1}{{\text{0}}^{-5}}\times 10 \\
 & \therefore {{n}_{{{\text{N}}_{\text{2}}}}}\text{ = 4}\times \text{1}{{\text{0}}^{-4}}\text{ mol} \\
\end{align}$
Therefore, the number of moles of nitrogen gas are equal to\[\text{ 4}\times \text{1}{{\text{0}}^{-4}}\text{ mol}\].

Hence, (A) is the correct option.

Note:
Henry's law is also written in terms of the mass of gas. If m is the mass of the gas dissolved per unit volume of as solvent and P is the pressure of the gas in equilibrium with the solution, then at a constant temperature the Henry law can be written as,
$\text{ m }\propto \text{ P or m = kP }$
Note that, if in any solution, the solute obeys Henry’s law within a certain range of concentration, the solvent obeys the Raoult’s law over the same range of concentration.