`PtCl_(4). 6H_(2)O` can exist as hydrated complex 1 molal aq. Solution has depression in freezing point of `3.72^(@)` . Assume `100%` ionisation and `K_(f)(H_(2)O) = 1.86^(@) mol^(-1)` kg , then complex is :

PtCl_(4).6H_(2)O can exist as a hydrated complex. 1 m aqueous solution has the depression in freezing point of 3.72^(@) . Assume 100% ionization and K_(f)(H_(2)O)=1.86^(@)mol^(-1) kg , then the complex is

PtCl_(4).6H_(2)O can exist as a hydrated complex. 1 m aqueous solution has the depression in freezing point of 3.72^(@) . Assume 100% ionization and K_(f)(H_(2)O)=1.86^(@)mol^(-1) kg , then the complex is

PtCl_(4).6H_(2)O can exist as a hydrated complex. 1 m aqueous solution has the depression in freezing point of 3.72^(@) . Assume 100% ionization and K_(f)(H_(2)O)=1.86^(@)mol^(-1) kg , then the complex is

PtCl_(4).6H_(2)O can exist as hydrated complex 1 molal aq.solution has depression in freezing point of 3.72^(@)C Assume 100% ionisation and K_(f)(H_(2)O=1.86^(@)mol^(-1))kg then complex is

PtCl_(4).6H_(2)O can exist as hydrated complex 1 molal aq.solution has depression in freezing point of 3.72^(@)C Assume 100% ionisation and K_(f)(H_(2)O=1.86^(@)mol^(-1))kg then complex is

PtCl_(4).6H_(2)O can exist as hydrated complex 1 molal aq.solution has depression in freezing point of 3.72^(@)C Assume 100% ionisation and K_(f)(H_(2)O=1.86^(@)Cmol^(-1)Kg) then complex is

The freezing point of a 0.05 m BaCl_(2) in water ( 100% ionisation) is about (K_(f)=1.86 Km^(-1)) :

Compound PdCl_(4). 6H_(2)O is a hydrated complex, 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex ( K_(f) for water = 1.86 K mol^(-1) )

Calculate depression of freezing point for 0.56 molal aq. Solution of KCl. (Given : K_f(H_(2)O) = 1.8 kg mol^(-1) ).