JEE Main20246 Apr 2024Morning ShiftChemistrySolutionsActual
Consider the dissociation of the weak acid ( HX ) as given below ( HX ( aq ) H ⁺( aq )+ X ⁻( aq ), Ka =1.2 10⁻⁵ ) ( [ K _ a . ) : dissociation constant (] ) The osmotic pressure of (0.03 M ) aqueous solution of ( HX ) at (300 ~K ) is _______ ( 10⁻² ) bar (nearest integer). [Given : ( R =0.083 ~L bar mol ⁻¹ ~K ⁻¹ )]
Correct answer
0
Step-by-step solution
aligned & HX H ⁺+ X ⁻ K _ a =1.2 10⁻⁵ & 0.03 M & 0.03- x x x & K _ a =1.2 10⁻⁵= x ^2 0.03- x & 0.03- x 0.03 ( ~K _ a is very small ) & x ^2 0.03 =1.2 10⁻⁵ & x =6 10⁻⁴ aligned Final solution : 0.03-x+x+x aligned & =0.03+x=0.03+6 10⁻⁴ & = (0.03+ (6 10⁻⁴ ) ) 0.083 300 & =76.19 10⁻² 76 10⁻² aligned