Question

The quantity of \[{\text{C}}{{\text{O}}_{\text{2}}}\] in $500{\text{ml}}$ of soda water when packed under ${\text{2}}{\text{.5atm C}}{{\text{O}}_{\text{2}}}$ pressure at $298{\text{K}}$ is $ - - - - - {\text{gm}}$ (Henry’s law constant $1.67 \times {10^8}{\text{pa at 298K}}$ )
A. $0.64$
B. $1.86$
C. $6.4$
D. $18.6$

Answer 1

Hint: Henry’s law states that the amount of gas dissolved in the liquid solution is directly proportional to the partial pressure of that gas at constant temperature. This law is valid at equilibrium, because this is the point when maximum gas is dissolved in the liquid solution.

Complete step by step answer:
Henry’s law is defined for gases. This law states that the amount of given gas that dissolves in a given liquid solution is directly proportional to the partial pressure of gas in equilibrium with the liquid at constant temperature.
According to the Henry’s law statement we can write,
${{\text{P}}_{{\text{gas}}}}{\text{ = }}{{\text{K}}_{\text{H}}}{\text{x}}$
Where ${{\text{P}}_{{\text{gas}}}}$ is the partial pressure of gas,
${{\text{K}}_{\text{H}}}$ is Henry’s law constant and
${\text{x}}$ is the mole dfraction of gas in the total mixture of gas and liquid.
In this problem we have given that
${{\text{P}}_{{\text{C}}{{\text{O}}_{\text{2}}}}} = 2.5{\text{atm = 2}}{\text{.5}} \times {\text{1}}{{\text{0}}^{\text{5}}}{\text{pa}}$
Volume of water=$500ml$ ($\therefore $ volume of ${\text{C}}{{\text{O}}_{\text{2}}}$ is negligible in the solution)
Therefore, mass of water=$500gm$
And we know that according to Henry’s law,
${{\text{P}}_{{\text{gas}}}}{\text{ = }}{{\text{K}}_{\text{H}}}{\text{x}}$
$ \Rightarrow {{\text{P}}_{{\text{C}}{{\text{O}}_{\text{2}}}}}{\text{ = }}{{\text{K}}_{\text{H}}}{\text{x}}$, ${\text{x}}$ is the mole dfraction of carbon dioxide gas.
Mole dfraction, ${\text{x}}$= $\dfrac{{{{\text{P}}_{{\text{C}}{{\text{O}}_{\text{2}}}}}}}{{{{\text{K}}_{\text{H}}}}} = \dfrac{{2.5 \times {{10}^5}}}{{1.67 \times {{10}^8}}} = 1.51 \times {10^{ - 3}}$
Mole dfraction, ${\text{x}}$= $\dfrac{{{\text{moles of C}}{{\text{O}}_{\text{2}}}}}{{{\text{moles of C}}{{\text{O}}_{\text{2}}} + {\text{ moles of }}{{\text{H}}_{\text{2}}}{\text{O}}}}$
Moles of water, ${{\text{H}}_{\text{2}}}{\text{O = }}\dfrac{{500}}{{18}} = 27.78moles$
Here the moles of carbon dioxide are negligible as compared to water, so some changes are made to above equation. The new equation will be:
${\text{x}}$= $\dfrac{{{\text{moles of C}}{{\text{O}}_{\text{2}}}}}{{{\text{ moles of }}{{\text{H}}_{\text{2}}}{\text{O}}}}$
${\text{moles of C}}{{\text{O}}_{\text{2}}} = {\text{x}} \times {\text{moles of }}{{\text{H}}_{\text{2}}}{\text{O}}$
${\text{moles of C}}{{\text{O}}_{\text{2}}} = {\text{x}} \times {\text{moles of }}{{\text{H}}_{\text{2}}}{\text{O}}$
${\text{moles of C}}{{\text{O}}_{\text{2}}} = {\text{1}}{\text{.51}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}} \times {\text{27}}{\text{.78 = 0}}{\text{.042moles}}$
So the obtained moles of carbon dioxide are $0.042$ .
Therefore the mass of carbon dioxide in the mixture is
Mass of carbon dioxide=${\text{moles of C}}{{\text{O}}_{\text{2}}} \times {\text{molar mass of C}}{{\text{O}}_{\text{2}}} = 0.042 \times 44$
Mass of carbon dioxide=$1.86gm$
Hence the weight of carbon dioxide gas present in the soda water is $1.86gm$ .

So, the correct answer is OptionB .

Note:
 In the formula of mole fraction we neglected the moles of carbon dioxide. We did this because carbon dioxide is present in a gaseous state and dissolved in liquid so it is clear that the mass of the gas will be very low as compared to mass of liquid that’s why the moles of carbon dioxide are neglected.