Answer

It is given that the solubility of ${H_2} {S}$ in water at STP is $0.195 {~m}$ i.e., 0.195 mol of ${H_2} {S}$ is dissolved in $1000 {~g}$ of water.

Moles of water $=\dfrac{1000 {~g}}{18 {~g} {~mol}^{-1}}$ $=55.56 {~mol}$

$ \text {Mole fraction of } H_2 S =\dfrac{\text { Moles of } {H_2} {S}}{\text { Moles of } {H_2} {S}+\text { Moles of water }} $

$ \hspace{3.9cm} =\dfrac{0.195}{0.195+55.56} =0.0035 $

At STP, pressure $(p)=0.987$ bar

According to Henry’s law : $ p={K_{H}} \chi$

$ \hspace{3.7cm} \Rightarrow {K_{H}}=\dfrac{p}{\chi}$

$ \hspace{3.7cm} =\dfrac{0.987}{0.0035} \hspace{0.5mm} bar $

$ \hspace{3.7cm} = 282 \hspace{0.5mm} bar $

Answer

It is given that :

${K_{H}}=1.67 \times 10^{8} {~Pa}$

$ p_{{CO_2}}=2.5 {~atm}=2.5 \times 1.01325 \times 10^{5} {~Pa} $

$\quad\quad\quad\quad\quad\quad\quad =2.533125 \times 10^{5} {~Pa}$

According to Henry’s law:

$p_{{CO_2}} ={K_{H}} \chi $

$\Rightarrow \chi =\dfrac{p_{{CO_2}}}{{~K_{H}}} $

$ \Rightarrow \chi =\dfrac{2.533125 \times 10^{5}}{1.67 \times 10^{8}}$

$\Rightarrow \chi=0.00152$

We can write,

$ \chi =\dfrac{n_{{CO_2}}}{n_{{CO_2}}+n_{{H_2} {O}}} \approx \dfrac{n_{{CO_2}}}{n_{{H_2} {O}}} $

[Since, $n_{{CO_2}}$ is negligible as compared to ${n_{{H_2} {O}}}$ ]

In $500 {~mL}$ of soda water, the volume of water $=500 {~mL}$ [Neglecting the amount of soda present]

We can write:

$500 {~mL}$ of water $=500 {~g}$ of water

$\quad\quad\quad\quad\quad\quad\quad=\dfrac{500}{18} {~mol}$ of water

$\quad\quad\quad\quad\quad\quad\quad=27.78 {~mol}$ of water

Now, $\quad\dfrac{n_{{CO_2}}}{n_{{H_2} {O}}}=\chi$

$\quad\quad \quad\dfrac{n_{{CO_2}}}{27.78}=0.00152$

$n_{{CO_2}}=0.042 {~mol}$

Hence, quantity of ${CO_2}$ in $500 {~mL}$ of soda water $=(0.042 \times 44) {g}$

$ \hspace{8.5cm}=1.848 {~g}$