Answer to Question #105081 in General Chemistry for Bright

Question

A mixture of 0.2 mole of alcohol A and 0.5 mole of alcohol B has a total Vapour pressure of 40 mmHg at 298k. If the mixture obeys Raoult's law, find the pure Vapour pressure of B at 298k given that the pure Vapour pressure of A is 20mmHg at 298k

Answer 1

p=p_A*x_A + p_B*x_B , where

• p - total vapour pressure of the mixture;
• p_A ,( p_B) -  pure vapour pressure of alcohol A, (B);
• x_A, x_B - mole fraction of A, (B);

p= 40 mmHg ;

p_A = 20 mmHg ;

n(A)=0.2 mole;

n(B)= 0.5 mole;

p_B - ?

x_A = n(A)/(n(A)+n(B)) = 0.2/0.7 = 0.2857 ;

x_B=n(B)/(n(A)+n(B)) = 0.5/0.7= 0.7143 ;

p_B= (p - p_A*x_A)/x_B = (40-20*0.2857)/0.7143 = 48 mmHg ;

Answer: The pure vapour pressure of B at 298k is 48mmHg