Answer

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

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

$=55.56 \mathrm{~mol}$

$$ =\frac{0.195}{0.195+55.56} $$

$$ =\frac{\text { Moles of } \mathrm{H_2} \mathrm{~S}}{\text { Moles of } \mathrm{H_2} \mathrm{~S}+\text { Moles of water }} $$

$=0.0035$

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

According to Henry’s law:

$p=\mathrm{K_\mathrm{H}} X$

$\Rightarrow \mathrm{K_\mathrm{H}}=\frac{p}{x}$

$$ =\frac{0.987}{0.0035} \text { bar } $$

$=282$ bar

Answer

It is given that:

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

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

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

According to Henry’s law:

$$ \begin{aligned} p_{\mathrm{CO_2}} & =\mathrm{K_\mathrm{H}} x \\ \Rightarrow x & =\frac{p_{\mathrm{CO_2}}}{\mathrm{~K_\mathrm{H}}} \\ & =\frac{2.533125 \times 10^{5}}{1.67 \times 10^{8}} \end{aligned} $$

$=0.00152$

We can write,

$$ x=\frac{n_{\mathrm{CO_2}}}{n_{\mathrm{CO_2}}+n_{\mathrm{H_2} \mathrm{O}}} \approx \frac{n_{\mathrm{CO_2}}}{n_{\mathrm{H_2} \mathrm{O}}} $$

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

In $500 \mathrm{~mL}$ of soda water, the volume of water $=500 \mathrm{~mL}$

[Neglecting the amount of soda present]

We can write:

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

$=\frac{500}{18} \mathrm{~mol}$ of water

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

Now, $\frac{n_{\mathrm{CO_2}}}{n_{\mathrm{H_2} \mathrm{O}}}=x$

$\frac{n_{\mathrm{CO_2}}}{27.78}=0.00152$

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

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

$=1.848 \mathrm{~g}$