Rate of effusion of LPG (a mixture of `n`-butane and propane) is 1.25 times that of `SO_(3)`. Hence, mole fraction of `n`-butane in LPG is

At 298K , the vapour pressure of pure liquid n-butane is 1823 torr and vapour pressure of pure n-pentane is 521 torr and form nearly an ideal solution. a. Find the total vapour pressure at 298 K of a liquid solution containing 10% n-butane and 90% n-pentane by weight, b. Find the mole fraction of n-butane in solution exerting a total vapour pressure of 760 torr. c. What is composition of vapours of two components (mole fraction in vapour state)?

At what temperature, the rate of effusion of N_(2) would be 1.625 times that of SO_(2) at 50^(@)C ?

LPG is a mixture of n-butane and iso-butane. The volume of oxygen needed to burn 1 kg of LPG at STP would be :

If LPG cylinder contains mixture of butane and isobutane, then the amount of oxygen that would be required for combustion of 1 kg of it will be

At what temperature, the rate of effusion of N_2 would be 1.625 times than the rate of SO_2 at 500^@C ?

If relative rates of substitution of 1^(@) and 2^(@) H are in the ratio 1 : 3.8, show that in the presence of light at 298 K, the chlorination of n-butane gives a mixture of 72% 2-chlorobutane and 28% 1-chlorobutane.