Home
Class 12
CHEMISTRY
The composition of the equilibrium mixtu...

The composition of the equilibrium mixture `(Cl_(2)//2Cl)` which is attained at `1200^(@)C`, is determined by measuring the rate of effusion through a pin-hole. It is observed that at 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as krypton effuse under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (atomic weight of Kr = 84 gm)

Text Solution

Verified by Experts

`(r_("mix"))/(r_("Kr"))=sqrt((M_("Kr"))/(M_("mix")))`
or `1.16 =sqrt((84)/(M))`
`(because` time of diffusion for both is same)
`therefore M =62.425`
`Cl_(2) hArr 2Cl`
For 1 0
`(1-alpha) 2alpha`
Total mole in mixutre `=1-alpha+2alpha=1+alpha`
`therefore 71xx(1-alpha)+2alphaxx35.5=(1+alpha)xx62.425`
`therefore alpha =0.137`
or 13.7%
Promotional Banner

Similar Questions

Explore conceptually related problems

The composition of the equilibrium mixture ( Cl_(2) 2Cl ) , which is attained at 1200^(@)C , is determined by measuring the rate of effusion through a pin hole. It is observed that a 1.80 mm Hg pressure, the mixture effuses 1.16 times as fact as krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (atomic weight of Kr is 84 ).

The composition of the equilibrium mixture ( Cl_(2) 2Cl ) , which is attained at 1200^(@)C , is determined by measuring the rate of effusion through a pin hole. It is observed that a 1.80 mm Hg pressure, the mixture effuses 1.16 times as fact as krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (atomic weight of Kr is 84 ).

The composition of the equilibrium mixture for the equilibrium Cl_(2)hArr2Cl at 1470^(@)K , may be determined by the rate of diffusion of mixture through a pin hole. It is found that at 1470^(@)K , the mixture diffuses 1.16 times as fast as krypton (83.8) diffuses under the same conditions. Calculate the % degree of dissociation of Cl2 at equilibrium.

At 800^(@)C, the following equilibrium is established as F_(2)(g)hArr2F(g) The cojmpositionof equilibrium may be determinded by measuring the rate of effusion of theh kmixture through a pin hole. It is found that at 800^(@)C and 1 atm mixture effuses 1.6 times as fast as SO_(2) effuse under the similar conditions. (At. mass of F = 19 ) what is the value of K_(p) (in atm) ?

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

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

1 L of a mixture of CO and CO_(2) , when passed through red hot tube containing charcoal, the volume becomes 1.6 L. all volumes are measured under the same conditions of temperature and pressure. Find the composition of the mixture.

The vapour of a substance effuses through a small hole at the rate 1.3 times faster than SO_(2) gas at 1 atm pressure and 500 K. The molecular weight of the gas is

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. If the vapour behaves ideally at 1000 K, determine the average translational kinetic energy of a molecule.