The statement "If 0.003 moles of a gas are dissolved in 900 g of water under a pressure of 1 atmosphere , 0.006 moles will be dissolved under a pressure of 2 atmosphere " illustrates
Let gas (A) present in air is dissolved in 20 moles of water at 298K and 20 atm pressure. The mole fraction of gas (A) in air is 0.2 and the Henry's law constant for solubility of gas (A) in water at 298K is 1×10^5 atm.The number of mole of gas (A) dissolved in water will be
For a solution of acetone in chloroform, Henry's law constant is 150 torr at a temperature of 300 K. (a) Calculate the vapour pressure of acetone when the mole fraction is 0.12. (b) Assuming that Henry's law is applicable over sufficient range of composition to make the calculation valid, calculate the composition at which Henry's law pressure of chloroform is equal to Henry's law pressure of acetone at 300 K. (Henry's law constant for chloroform is 175 torr.)
The vapour pressure of water is 17.54mm Hg at 293 K. Calculate vapour pressure of 0.5 molal solution of a solute in it.
At 760 torr pressure and 20^(@)C tempreature , 1 L of water dissolves 0.04 gm of pure oxygen or 0.02 gm of pure nitrogen. Assuming that dry air is compound of 20% oxygen and 80% nitrogen (by volume), the masses (in g/L) of oxygen and nitrogen dissolved by 1 L of water at 20^(@)C exposed to air at a total pressur of 706 torr are respectively :
NARENDRA AWASTHI-DILUTE SOLUTION-Level 3 - Match The Column
