Home
Class 12
CHEMISTRY
The density of vapours of a substance at...

The density of vapours of a substance at 1 atm and 500 K is `0.3 kg m^(-3)`. The vapours effuse 0.4216 times faster than `O_(2)` through a pin hole under identical conditions. If R= 0.08 litre atm `K^(-1) mol^(-1)`. The molar volume of gas is `a xx 10^(2)` litre. The value of a is _____________.

Text Solution

Verified by Experts

The correct Answer is:
6
Let the Mol. Wt. of substance be MO g/mol `:. ("Rate of effusion of substance")/("Rate of effusion of " O_(2)) = sqrt((32)/(M_(0)))`
`rArr (0.4216)^(2) = (32)/(M_(o)) rArr M_(o) = 180g//mol`
As density of substance = 0.3 g/L `:.` Vol. of 1 mole `=(180g//mol)/(0.3g//L) = 600L`
or, `6 xx 10^(2)L`, So, a = 6
Promotional Banner

Similar Questions

Explore conceptually related problems

The density of vapour of a substance (X) at 1 atm pressure and 500 K is 0.8 kg//m^(3) . The vapour effuse through a small hole at a rate of 4//5 times slower than oxygen under the same condition. What is the compressibility factor (z) of the vapour ?

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg m^(-3) . The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. ( a ) Determine ( i ) the molecular weight, ( ii ) the molar volume ( iii ) the compression factor( Z ) of the vapour, and ( iv ) which forces among the gas molecules are dominating, the attractive or the repulsive? ( b ) If the vapour behaves ideally at 100 K , determine the average translational kinetic energy of a molecule.

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg m^(-3) . The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. Determine, (a) molecular weight (b) molar volume (c) compression factor (Z) of the vapour and (d) which forces among the gas molecules are dominating, the attractive or the repulsive?

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg m^(-3) . The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. If the vapour behaves ideally at 1000 K, determine the average translational kinetic energy of a molecule.

The density of the vapour of a substance at 1 atm and 500 K is 0.36 kgm^(-2) . If molar mass of gas is 18 g mol^(-1) the molar volume of gas is 5 xx 10^(a) m^(3)//mol What is the value of a ?

If density of vapours of a substance of molar mass 18 gm at 1 atm pressure and 500K is 0.36 kg m^(-3) , then calculate the value of Z for the vapours. (Take R = 0.082 L atm mole^(-1)K^(-1))