Question

Concentration of $CO _{2}$ (in mole fraction) in fat when partial pressure of $CO _{2}$ is $55 \,kPa$ at $25^{\circ} C$, is (Henry's law constant of $CO _{2}=8.6 \times 10^{4}$ torr $)$

• A. $2.6 \times 10^{-4}$
• B. $4.80 \times 10^{-3}$
• C. $1.2 \times 10^{4}$
• D. $8.2 \times 10^{-4}$

Answer 1

Answer: (B)

$K _{ H }$ (Henry's law constant) is in pressure unit, hence we use relation, Concentration $\times K _{ H }=$ Pressure

$\therefore $ Concentration $=\frac{\text { Pressure of } C O _{2}}{K_{H}}$

$p_{ CO _{2}}=55\, kPa =55 \times 10^{3} Pa$

(Pa unit is to be converted into torr)

$1 \,atm =760\, torr =1.01325 \times 10^{5} Pa $

$\therefore 55 \times 10^{3} Pa =\frac{760 \times 55 \times 10^{3}}{1.01325 \times 10^{5}} torr =412.53\, torr$

$\therefore \chi_{ CO _{2}}$ (Mole fraction) $=\frac{412.53 \,torr }{8.6 \times 10^{4} torr } $

$=4.80 \times 10^{-3}$