mole fraction of solute. The vapour...
Question  # 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

JEE/Engineering Exams
Chemistry
Solution 129 4.0 (1 ratings)  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} ) Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free