Question

When 3.2 g of Sulphur is vaporized at 450oC and 723 mm pressure, the vapors occupy a volume of 780 ml. What is the molecular formula of sulphur vapours under these conditions? Calculate the vapor density also.

Answer 1

The Reaction would be:

Total moles = moles of sulphur + moles of Suphur vapor.

= 3.2/32 + n

From ideal gas eqn. we get,

(723/760)*(780/1000) = (n+0.1)*0.0821*723

(n+0.1) = [(723/760)*(780/1000)]/[(0.0821*723)] = 0.0125

as n is coming negative and hence the moles can’t be nagtive the Reaction is not feasible here.