Air contains O2 and N2 in the ratio 0.2 : 0.8. If Henry law constant for O2 and N2 are 3.3 × 107 torr and 6.6 × 107 torr respectively, then the ratio of mole fractions of O2 to N2 dissolved in water at 1 bar pressure is

According to Henry's law,P=KHX⇒X=PKH XO2XN2=PO2PN2×KH(N2)KH(O2)=0.20.8×6.6×1073.3×107=12

Air contains O2 and N2 in the ratio 0.2 : 0.8. If Henry law constant for O2 and N2 are 3.3 × 107 torr and 6.6 × 107 torr respectively, then the ratio of mole fractions of O2 to N2 dissolved in water at 1 bar pressure is

According to Henry's law,P=KHX⇒X=PKH XO2XN2=PO2PN2×KH(N2)KH(O2)=0.20.8×6.6×1073.3×107=12

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Air contains O₂and N₂ in the ratio 0.2 : 0.8. If Henry law constant for O₂ and N₂ are 3.3 × 10⁷ torr and 6.6 × 10⁷torr respectively, then the ratio of mole fractions of O₂ to N₂ dissolved in water at 1 bar pressure is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

1:1

2:1

1:2

1:3

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$According to Henry' s law, P = K_{H} X \Rightarrow X = \frac{P}{K_{H}} \frac{X_{O_{2}}}{X_{N_{2}}} = \frac{P_{O_{2}}}{P_{N_{2}}} \times \frac{K_{H} (N_{2})}{K_{H} (O_{2})} = \frac{0.2}{0.8} \times \frac{6.6 \times 10^{7}}{3.3 \times 10^{7}} = \frac{1}{2}$