Home
Class 12
CHEMISTRY
ne solution becomes 0.9. The [JEE 2008, ...

ne solution becomes 0.9. The [JEE 2008, 3/1631 wat ine mole fraction of water in the solution bec Surution is TA) 380.4 K (B) 376.2 K (C) 375.5 K (D) 354.7 K The Henry's law constant for the solubility of Ne gas in water at 298 K is 1.0 x 105 atm The fraction of N2 in air is 0.8. The number of moles of Ne from air dissolved in 10 moles of water of pag atm. The mole and 5 atm pressure is : (A) 4 x 10 (B) 4.0 x 10-5 (C) 5.0 x 10-4 (D) 4.0 x 10-6 The freezing point (in ºC) of a solution containing 0.1 g of K3[Fe(CN K = 1.86 K kg mol-1) is: [JEE 2009, 3/160]

Promotional Banner

Similar Questions

Explore conceptually related problems

The 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 0.8. The number of moles of N_2 from air dissolved in 10 mole of water at 298 K and 5 atm pressure is:

The Henry's law constant for the solubility of N_(2) gas in water at 298K is 1.0 xx 10^(5) atm. The mole fraction of N_(2) in air is 0.8 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298K and 5 atm pressure is

The Henry's law constant for the solubility of N_(2) gas in water at 298K is 1.0 xx 10^(5) atm. The mole fraction of N_(2) in air is 0.8 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298K and 5 atm pressure is

The Henry's law constant for the solubility of N_(2) gas in water at 298K is 1.0 xx 10^(5) atm. The mole fraction of N_(2) in air is 0.8 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298K and 5 atm pressure is

The Henry's law constant fo the solubility of N_(2) gas in water at 298 K is 1.0xx10^(5) atm. The mole fraction of N_(2) in air is 0.8. The number of moles of N_(2) from air dissolved in 10 moles of water at 298K and 5 atm pressue is