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Henry’s law constant for $C O_{2}$ in water is $1.67 \times 10^{8} P a$ at 298K. The quantity (in grams) of $C O_{2}$ in 500mL of soda water when packed under 2.5 atm $C O_{2}$ pressure at 298K is________

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Detailed Solution

$P_{C O_{2}} = K_{H} χ_{C O_{2}}$

$2.5 \times 1.0 1325 \times 10^{5} = 1.67 \times 10^{8} χ_{C O_{2}}$

$χ_{C O_{2}} = 1.52 \times 10^{- 3}$

$1.52 \times 10^{- 3} = \frac{n_{C O_{2}}}{n_{H_{2} O}} [n_{H_{2} O} = \frac{500}{18} = 27.78]$

$n_{C O_{2}} = 1.52 \times 10^{- 3} \times 27.78$

$= 0.042 m o l e s$

$w t o f C O_{2} = 44 \times 0.042 = 1.848 g$

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