Henry's law constant for $C O_{2}$ in water is $2.5 \times 10^{8}$ Pa at 298 K. Calculate milli mole of $C O_{2}$ dissolved in 162 g water at 2.5 atm pressure at 298 K.

$[T a k e 1 a t m = 10^{5} N / m^{2}]$

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Solution

The correct option is C 9
$P_{C O_{2}} = K_{H} . χ_{C O_{2}} = K_{H} \times \frac{n_{C O_{2}}}{n_{C O_{2}} + n_{w a t e r}}$

$= K_{H} \frac{n_{C O_{2}}}{n_{w a t e r}}$

$2.5 \times 10^{5} = 2.5 \times 10^{8} \times \frac{n_{C O_{2}}}{162 / 18}$

$n_{C O_{2}} = 9 \times 10^{- 3} m o l e s$ or $9 m i l l i m o l e s$

Hence, the correct answer is option $A$ .

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Henry's law constant for CO2 in water is 2.5×108 Pa at 298 K. Calculate milli mole of CO2 dissolved in 162 g water at 2.5 atm pressure at 298 K. [Take1atm=105N/m2]

The correct option is C 9PCO2=KH.χCO2=KH×nCO2nCO2+nwater =KHnCO2nwater 2.5×105=2.5×108×nCO2162/18 nCO2=9×10−3 moles or 9 millimolesHence, the correct answer is option A.

Henry's law constant for $C O_{2}$ in water is $2.5 \times 10^{8}$ Pa at 298 K. Calculate milli mole of $C O_{2}$ dissolved in 162 g water at 2.5 atm pressure at 298 K.

$[T a k e 1 a t m = 10^{5} N / m^{2}]$