Home
Class 11
CHEMISTRY
Rate of diffusion of LPG (a mixture of n...

Rate of diffusion 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

A

`0.752`

B

`0.256`

C

`0.514`

D

`0.667`

Text Solution

Verified by Experts

The correct Answer is:
C
Since, `(r_(SO_3))/(r_("mix"))= sqrt((M_("mix"))/(M_("mix"))) implies(1)/(1.25)=sqrt((M_("mix"))/(80)) implies M_("mix")= 51.2`
Since, `M_("mix")= x_("butane")M_("butane")+x_("propane")M_("propane")`, Therefore, `51.2= 58x_("butane")+44(1-x_("butane"))`
`implies 7.2 = 14x_("butane") implies x_("butane")= 0.514`.
Promotional Banner

Similar Questions

Explore conceptually related problems

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

The vapour density of a mixture containing NO_(2) and N_(2)O_(4) is 38.3 at 27^(@)C . Calculate the mole of NO_(2) in 100mole of mixture

For a chemicai reaction (A) rarr (B) it is found that the rate of the reaction guadruples when the concentration of (A) is doubled.The rate expression for the reaction is rate =(k)[(A)] (n) , where the value of (n) is a)1 b)2 c)0 d)3

Explain why chlorination of n-butane in presence of light at 298 K gives a mixture of 72 % 2-chlorobutane and 28 % of 1- chlorobutane.

Rate of formation of SO_(3) in the following reaction 2SO_(2) + O_(2) to 2SO_(3) is 100g "min"^(-1) . Hence rate of disappearance of O_(2) is: