Question

Let the molality of the solution ( =m )
Now the solution contains ( mathrm{m} ) moles of solute per ( 1000 mathrm{gm} ) of benzene.
Vapour pressure of pure benzene, ( P^{0}=639.7 mathrm{mm} )
Vapour pressure of solution, ( mathrm{P}=631.9 mathrm{mm} )
Moles of benzene ( (mathrm{mol} text { wt. } 78) N=frac{1000}{78} )
Moles of solute, ( n=? )
Substituting these values in the Raoult's equations
( frac{P^{0}-P}{P^{0}}=frac{n}{N} )
( frac{639.7-631.9}{639.7}=frac{n times 78}{1000} )
(Or) ( frac{7.8}{639.7}=frac{78 n}{1000} )
( therefore n=frac{1000 times 7.8}{78 times 639.7} )
( Rightarrow 0.156 )
Hence molality of solution ( =0.156 mathrm{mole} / mathrm{kg} )

# mole fraction of solute. The vapour pressure of pure benzene at 25° C is 639.7 mm of Hg and the vapour pressure of a soluti of a solute in C H at the same temperature is 631.7 mm of Hg. Calculate molality of solution

Solution