The quantity of ${CO}_2$ in 500ml of soda water when packed under 2.5 atm ${CO}_2$ pressure at 298K is ____ gm ( Henry's law constant $1.67\times {10}^8$ pa at 298K)

The correct option is A 1.86

According to Henry's law, $p_{C{O}_2}=K_{H}x$

Here $x$ is the mole fraction of carbon dioxide in the carbon dioxide water mixture.

$p_{CO2}=2.5atm=2.5\times 1.01325\times {10}^{5}Pa=2.533125\times {10}^{5}Pa$

$x=\frac{molesofC{O}_2}{molesofC{O}_2+molesof{H}_{2}O}$

Hence,

According to Henry's law, $p_{C{O}_2}=K_{H}x$

$\therefore x=molefraction=\frac{p^{}C{O}_2}{K_H}=\frac{2.533125\times {10}^5}{1.67\times {10}^8}=1.51\times {10}^{-3}$

Here moles of $C{O}_2$ is negligible in the denominator as compared to moles of $H_{2}O$

500 ml of water = 500g of water = $\frac{500}{18}$= 27.78 moles of water.

So, moles of $\frac{C{O}_2}{27.78}=1.51\times {10}^{-3}$ or moles of $C{O}_2=0.042moles.$

Hence the amount of $C{O}_2$ in 500 ml of soda water is $0.042\times 44=1.86g$

Hence, the correct option is $B$