For the solution of the gases ` W, X, Y ` and z in water at 298 K , the Henrys law constants (`K _ H `) are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is :
Cacluate the amount of CO_(2) dissolved at 4 atm in 1 "dm"^(3) of water at 298 K . The Henry's law constant for CO_(2) at 298K is 1.67 k bar .
At what partial pressure , oxygen will have a solubility of 0.06 gL^(-1) in water at 298 K ? Henry's law constant (K_(H)) of O_(2) in water at 303 K is 46.82 k bar .(Assume the density of the solution to be the same as that of water).
The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20%:79% by volume at 298K . The water is in equilibrium with air at a pressure of 10atm At 298 K if Henry's law constants for oxygen and nitrogen at 298 K are 3.30 xx 10^(7)mm and 6.51xx10^(7)mm , respectively, calculate the composition of these gases in water.
The values of the Henry's law constant of Ar, CO_2 , CH_4 ,and O_2 in water at 25^@C are 40.30,1.67,0.41 and 34.86 kbar, respectively, The order of their solubility in water at the same temperature and pressure is
Four gases alpha, beta, gamma and delta have K_H values 50kbar, 20 kbar, 2xx10^(-3) kbar and 2 kbar respectively. then ?
The Henry’s law constant of a gas is 6.7×10^-4 mol/(L bar). Its solubility when the partial pressure of the gas at 298K is 0.65 bar is
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