Concentration of CO 2 (in mole fraction) in fat when partial pressure of CO 2 is 55 kPa at 25° C, is (Henry's law constant of CO 2=8.6 × 104 torr )

Question

Concentration of CO 2 (in mole fraction) in fat when partial pressure of CO 2 is 55 kPa at 25° C, is (Henry's law constant of CO 2=8.6 × 104 torr ) (A) 2.6 × 10-4 (B) 4.80 × 10-3 (C) 1.2 × 104 (D) 8.2 × 10-4

Tardigrade Admin · Accepted Answer

K H (Henry's law constant) is in pressure unit, hence we use relation, Concentration × K H = Pressure ∴ Concentration =( text Pressure of C O 2/KH) p CO 2=55 kPa =55 × 103 Pa (Pa unit is to be converted into torr) 1 atm =760 torr =1.01325 × 105 Pa ∴ 55 × 103 Pa =(760 × 55 × 103/1.01325 × 105) torr =412.53 torr ∴ chi CO 2 (Mole fraction) =(412.53 torr /8.6 × 104 torr ) =4.80 × 10-3

Q. Concentration of $C O_{2}$ (in mole fraction) in fat when partial pressure of $C O_{2}$ is $55 k P a$ at $2 5^{\circ} C$ , is (Henry's law constant of $C O_{2} = 8.6 \times 1 0^{4}$ torr $)$

$2.6 \times 1 0^{- 4}$

$4.80 \times 1 0^{- 3}$

$1.2 \times 1 0^{4}$

$8.2 \times 1 0^{- 4}$

Q. Concentration of CO2​ (in mole fraction) in fat when partial pressure of CO2​ is 55kPa at 25∘C, is (Henry's law constant of CO2​=8.6×104 torr )

2.6×10−4

4.80×10−3

1.2×104

8.2×10−4

Q. Concentration of $C O_{2}$ (in mole fraction) in fat when partial pressure of $C O_{2}$ is $55 k P a$ at $2 5^{\circ} C$ , is (Henry's law constant of $C O_{2} = 8.6 \times 1 0^{4}$ torr $)$

$2.6 \times 1 0^{- 4}$

$4.80 \times 1 0^{- 3}$

$1.2 \times 1 0^{4}$

$8.2 \times 1 0^{- 4}$