The time for a certain volume of a gas `A` to diffuse through a small hole is 2 minute If takes 5.65 minute for oxygen to diffuse under similar conditions Find the molecualr weight of `A` .
The Graham's law states that ''at constant pressure and temperature the rate of diffusion or effusion of a gas is inversely proportional to the squar root of its density Rate of diffusion prop (1)/(sqrt(d)) If r_(1) and r_(2) represent the rates of diffusion of two gases and d_(1) and d_(2) are their respective densities, then r_(1)/(r_(2))=sqrt((d_(2))/(d_(1))) r_(1)/(r_(2)) =sqrt((M_(2))/(M_(1))) xx P_(1)/(P_(2)) (V_(1)xxt_(2))/(V_(2)xxt_(1)) = sqrt((d_(2))/(d_(1))) = sqrt((M_(2))/(M_(1))) V prop n (where n is no of moles) V_(1) prop n_(1) and V_(2) prop n_(2) The time taken for a certain volume of gas X to diffuse through a small hole is 2 minutes It takes 5.65 minutes for oxygen to diffuse under the simillar conditions The molecular weight of X is .
50 mL of hydrogen diffuses through a small hole from vessel in 20 minutes time. Time taken for 4 mL of oxygen to diffuse out under similar conditions will be
50 mL of hydrogen diffuse through a small hole from a vessel in 20 mintues time. Time taken for 40 ml of oxygen to diffuse out under similar conditions will be :
The time taken for a certain volume of gas to diffuse through a small hole was 2 min . Under similar conditions an equal volume of oxygen took 5.65 minute to pass. The molecular mass of the gas is
A gaseous mixture of O_(2) and an unknown gas 'X' containing 20 mole % of X diffused through a small hole in 245 seconds while O_(2) takes 220 seconds to diffuse through the same hole under similar conditions. Calculate the molecular mass of X.
50 mL of hydrogen takes 10 minutes to diffuse out of a vessel. How long will 40 mL of oxygen take to diffuse out under similar conditions ?
