Home
Class 12
CHEMISTRY
Henry 's law constant for the solubility...

Henry 's law constant for the solubility of `N_(2)` gas in water at `298 K` is `1.0 xx 10^(5) atm`. The mole fraction of `N_(2)` in air is `0.6`. The number of moles of `N_(2)` from air dissolved in 10 moles of water at `298 K`and `5 atm` pressure is
a. `3.0 xx 10^(-4)` b. `4.0 xx 10^(-5)` c. `5.0 xx 10^(-4)` d.`6.0 xx 10^(-6)`

Text Solution

Verified by Experts

a. Partial pressure of `N^(2)` in air `(p_(N_2)) = P_("total") xx (chi_(N_2))("in air")`
`=5 xx 0.6`
`P_(N_2)`("in air" ) `= K_H xx chi_(N_2)("in" (H_(2)O))`
`5 xx 0.6 = 1.0 xx 10^(5) xx chi^(N_(2))("in" H_(2)O)`
`chi_(N_(2))` in 10 moles of water `=(5 xx 0.6)/(1.0 xx 10^(5)) = 3.0 xx 10^(-5)`
`chi_(N_(2))= n_(N_(2))/(n_(N_(2)) + n_(H2O))`
`3.0 xx 10^(-5) = n_(N_(2))/(n_(N_(2)) + 10)`
`n_(N_(2)) xx 3 xx10^(-5) xx 10 = n_(N_(2))`
`3 xx 10^(-4) = n_(N_(2)) (1-3 xx 10^(-5)) [1-3 xx 10^(-5) = 1]`
`:. n^(N_(2)) = 3 xx 10^(-4)`
Promotional Banner

Similar Questions

Explore conceptually related problems

When 0.1 mole of glucose is dissolved in 10 mole of water, the vapour pressure of water is

The solubility product of HgCl_(2) at room temperature is 4.0 xx 10^(-5) . The concentration of Cl^(-) ion in the saturated solution will be

Henry's law constant for the molality of methane in benzene of 298 K is 4.27 xx 10^(5) mm Hg. Calculate the solubility of methane in benzene of 298 K under 760 mm Hg.

Solubility product of PbCl_(2) at 298 K is 1.0xx10^(-6) . At this temperature solubility of PbCl_(2) in moles per litre is :

Henry law constant for oxygen dissolved in water is 4.34 xx 10^(4) atm at 25^(@) C . If the partial pressure of oxygen in air is 0.4 atm.Calculate the concentration ( in moles per litre) of the dissolved oxygen in equilbrium with air at 25^@C .

Solubility of an AB_(2) type electrolyte is 5.0xx10^(-5) mol L^(-1). K_(sp) for the electrolyte AB_(2) is :