Henry's law constant for ${C O}_{2}$ in water is $1.67 \times 10^{8}$ Pa at 298K. Calculate the quantity of ${C O}_{2}$ in 500 ml of soda water when packed under 2.5 atm ${C O}_{2}$ pressure at 298 K.

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Solution

According to Henry's law :

$p = K_{H} x .$

$P {C O}_{2} = 2.25 a t m = 2.5 \times 1.01325 \times 10^{5} P a = 2.533 \times 10^{5} P a .$

$∴ x = \frac{2.533 \times 10^{5}}{1.67 \times 10^{8}} x = 0.00152.$

$X_{{C O}_{2}} = \frac{{n C O}_{2}}{{n C O}_{2} + {n H}_{2} O} 500 H_{2} O = 500 H_{2} O$

moles of $H_{2} O = \frac{500}{18} = 27.78$

$0.00152 = \frac{{n C O}_{2}}{{n C O}_{2} + 27.78}$

${n C O}_{2} = 0.042$

moles of ${C O}_{2} = m o l e s \times m o l a r m a s s = 0.042 \times 44 = 1.848.$

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Henry's law constant for CO2 in water is 1.67×108Pa at 298K. Calculate the quantity of CO2 in 500 ml of soda water when packed under 2.5 atm CO2 pressure at 298 K.

According to Henry's law : p=KHx. PCO2=2.25atm=2.5×1.01325×105Pa=2.533×105Pa. ∴x=2.533×1051.67×108x=0.00152. XCO2=nCO2nCO2+nH2O500H2O=500H2O moles of H2O=50018=27.78 0.00152=nCO2nCO2+27.78 nCO2=0.042 moles of CO2=moles×molarmass=0.042×44=1.848.

Henry's law constant for ${C O}_{2}$ in water is $1.67 \times 10^{8}$ Pa at 298K. Calculate the quantity of ${C O}_{2}$ in 500 ml of soda water when packed under 2.5 atm ${C O}_{2}$ pressure at 298 K.