Henry law constant for co2 dissolving …

It is given that:
KH = {tex}1.67\times10^8Pa{/tex}
{tex}{P_{c{o_2}}}{/tex} = 2.5 atm = {tex}2.5\times1.01325\times10^5Pa{/tex}
{tex}=2.533125\times10^5Pa{/tex}
According to Henry's law:
{tex}P_{CO_2}=K_{H\;}\;x{/tex}
{tex}x = \frac{{Pc{o_2}}}{{{K_H}}}{/tex}
{tex}\frac{{2.533125 \times {{10}^5}}}{{1.67 \times {{10}^8}}}{/tex}
= 0.00152
We can write, {tex}x = \frac{{{n_{co}}_{_2}}}{{{n_{c{o_2}}} + {n_{{H_2}O}}}} \approx \frac{{{n_{c{o_2}}}}}{{{n_{{H_2}O}}}}{/tex}
[Since, {tex}{n_{c{o_2}}}{/tex}is negligible as compared to {tex}{n_{{H_2}O}}{/tex}]
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
={tex}\frac{{500}}{{18}}{/tex} mol of water
= 27.78 mol of water
Now, {tex}\frac{{{n_{c{o_2}}}}}{{{n_{{H_2}O}}}} = x{/tex}
{tex}\frac{{{n_{c{o_2}}}}}{{27.78}} = 0.00152{/tex}
{tex}{n_{c{o_2}}} = 0.042\,mol{/tex}
Hence, quantity of CO₂ in 500 mL of soda water = {tex}(0.042 \times 44)g{/tex}
= 1.848 g