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

The Henry's law constant for the solubility of nitrogen gas in water at 298 K is `1.0 xx 10^5 a t m`. The mole fraction of `N_2` in air is `0.8`. The number of moles of `N_2` from air dissolved in 10 moles fo water at 298 K and 5 atm. Pressure is :

A

`4.0 xx 10^(-4) atm`

B

`4.0 xx 10^(-5) atm`

C

`5.0 xx 10^(-4) atm`

D

`4.0 xx 10^(-6) atm`

Text Solution

Verified by Experts

The correct Answer is:
A
`P_(N2)=5 xx 0.8 =4`
Applying Henry's law, we get
`P_(N2)=K_(H) xx chi_(N_(2))` or `4=10^(5) xx chi_(N_(2))`
chi_(N_(2))=4 xx 10^(-5)`
`n_("nitrogen")/(n_("nitrogen")+n_(H_(2)O))=4 xx 10^(-5)`
or `n_("nitrogen")/(n_("nitrogen")+10)=4 xx 10^(-5)`
or `n_("nitrogen")=4 xx 10^(-4)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

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)

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

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.

The solubility of a gas in water at 300 K under a pressure of 100 atmospheres is 4xx10^(-3) "kg L"^(-1) . Therefore, the mass of the gas in kg dissolved in 250 mL of water under a pressure of 250 atmospheres at 300 K 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 .

Henry's law constant for oxygen and nitrogen dissolved in water at 298 K are 2.0 xx 10^(9) Pa and 5.0 xx 10^(9) Pa , respectively . A sample of water at a temperature just above 273 K was equilibrated with air (20% oxygen and 80% nitrogen ) at 1 atm. The dissolved gas was separated from a sample of this water and the dried. Determine the composition of this gas.

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