At 730° C the vapor density of iodine under a pressure of 1 atm is 97.45 % of the theoretical, on the assumption that the molecule of gaseous iodine is diatomic. What is the degree of dissociation of I2 ?

Question

At 730° C the vapor density of iodine under a pressure of 1 atm is 97.45 % of the theoretical, on the assumption that the molecule of gaseous iodine is diatomic. What is the degree of dissociation of I2 ? (A) 0.0131 (B) 0.0262 (C) 0.082 (D) 0.0324

Tardigrade Admin · Accepted Answer

van't Hoff factor i=( text Theoretical value of mol. wt. / text Observed value of mol. wt. ) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/2d83eec5c4a535c02330482c2bfc759e-.png" /> Total number of moles at equilibrium =1-α+2 α=1+α i=(1+α/1) ; (254/247.52)=1+α or 247.52 α=254-247.52=6.48. α=(6.48/247.52)=0.0262

Q. At $73 0^{\circ} C$ the vapor density of iodine under a pressure of $1$ atm is $97.45%$ of the theoretical, on the assumption that the molecule of gaseous iodine is diatomic. What is the degree of dissociation of $I_{2}$ ?

0.0131

0.0262

0.082

0.0324

van't Hoff factor

$i = \frac{Theoretical value of mol. wt.}{Observed value of mol. wt.}$

Total number of moles at equilibrium $= 1 - α + 2 α = 1 + α$

$i = \frac{1 + α}{1}; \frac{254}{247.52} = 1 + α$

or $247.52 α = 254 - 247.52 = 6.48.$

$α = \frac{6.48}{247.52} = 0.0262$

Q. At 730∘C the vapor density of iodine under a pressure of 1 atm is 97.45% of the theoretical, on the assumption that the molecule of gaseous iodine is diatomic. What is the degree of dissociation of I2​ ?