Heny’s law constant for \mathrm{CO}_{2} in water is 1.67 \times 10_{8} \mathrm{~Pa} at 298 \mathrm{~K}. Calculate the quantity of \mathrm{CO}_{2} in 500 \mathrm{~mL} of soda water when packed under 2.5 \mathrm{~atm} \mathrm{CO}_{2} pressure at 298 \mathrm{~K}.
Heny’s law constant for \mathrm{CO}_{2} in water is 1.67 \times 10_{8} \mathrm{~Pa} at 298 \mathrm{~K}. Calculate the quantity of \mathrm{CO}_{2} in 500 \mathrm{~mL} of soda water when packed under 2.5 \mathrm{~atm} \mathrm{CO}_{2} pressure at 298 \mathrm{~K}.

Solution 7:

    \[\mathrm{KH}=1.67 \times 10_{8} \mathrm{~Pa}\]

    \[\begin{array}{l} \hline P_{A}^{o}=450 \mathrm{~mm}, P_{B}^{o}=700 \mathrm{~mm}, P_{\text {total }}=600 \mathrm{~mm} \\ \text { As Ptotal }=\mathrm{PA}+\mathrm{PB} \\ =X_{A} P_{A}^{o}+\left(1-X_{A}\right) P_{B}^{o} \\ =P_{B}^{o}+\left(P_{A}^{o}-P_{B}^{o}\right) X_{A} \\ \Rightarrow 600=700+(450-700) X_{A} \\ \text { Or X }_{\mathrm{A}}=0.40 \\ \therefore x_{B}=1-X_{A}=1-1.40=0.60 \\ \therefore P_{A}-x_{A} P_{A}^{o}=0.40 \times 450=180 \mathrm{~mm} \\ \therefore=x_{B} P_{A}^{o}=0.60 \times 700=420 \mathrm{~mm} \end{array}\]

Mole fraction of \mathrm{A} in vapour phase

    \[=\frac{P_{A}}{P_{A}+P_{B}}=\frac{180}{180+420}=0.30\]

And, mole fraction of \mathrm{B} in vapour phase =1-0.30=0.70
Vapour pressure of pure water at 298 \mathrm{~K} is 23.8 \mathrm{~mm} \mathrm{Hg}. 50 \mathrm{~g} of urea \left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right) is dissolved in $850