AMU Engineering 2018 solved paper Solved Paper

The solubility of gas in liquid at particular temperature is given by Henry's law.
i.e. mA=KHpA
where, mA= mass of gas dissolved in a unit volume of solvent
pA= pressure of the gas in equilibrium
Kmathrm‌H= Henry's law constant
On putting the given values in above expression 1.38×10−3mol∕L=KH×1atm
KH=

1.38×10−3mol∕L

1atm

At 0.21 atm, mA will be calculated as mA=

1.38×10−3mol∕L

1atm

×0.21atm
mA=2.9×10−4mol∕L