Question bank/JEE Main/Chemistry

JEE MainChemistryModeratePhysical Chemistry

The Henry's law constant for the solubility of N2 gas in water
at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The

number of moles of N2 from air dissolved in 10 moles of water of 298 K

and 5 atm pressure is:

N2 xSl dh ty esa ? fy, gsujh fu;e fLFkjkad (Henry's law

constant) dk eku 298 K ij 1.0 × 105 atm gSA N2 izHkkt

mifLFkr N2

Show answer and solution
PN2=KH×XN2P_{N_{2}} = K_{H} \times X_{N_{2}}
eksy

In 10 mole solubility is eksy

Source: Grass-G / CH-27 Chemistry Paper 2 15 Jan.docx

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