In aqueous solution of `1times10^(-3)` molal `K_(x)[Fe(CN)_(6)]` depression in freezing point is `7.2times10^(-3)K`. Determine sum of primary and secondary valency of complex `(K_(f)` of `H_(2)O=1.8(K.kg)/(mol))`. (Assume that `%` ionisation of complex is `100%`)

1 xx 10^(-3) m solution of Pt(NH_(3))_(4)Cl_(4) in H_(2)O shows depression in freezing point of 0.0054^(@)C . The formula of the compound will be [Given K_(f) (H_(2)O) = 1.86^(@)C m^(-1) ]

An aqueous solution of 0.1 molal concentration of sucrose should have freezing point (K_(f)=1.86K mol^(-1)kg)

The freezing point depression of 0.001 m K_(x) [Fe(CN)_(6)] is 7.10xx10^(-3) K . Determine the value of x. Given, K_(f)=1.86 K kg "mol"^(-1) for water.

The freezing point depression of 0.001 m K_(x) [Fe(CN)_(6)] is 7.10xx10^(-3) K . Determine the value of x. Given, K_(f)=1.86 K kg "mol"^(-1) for water.

0.01 molal aqueous solution of K_(3)[Fe(CN)_(6)] freezes at -0.062^(@)C . Calculate percentage dissociation (k_(f)=1.86)

Molal depression constant for a solvent is 4.0 kg mol^(-1) . The depression in the freezing point of the solvent for 0.03 mol kg^(-1) solution of K_(2)SO_(4) is : (Assume complete dissociation of the electrolyte)

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

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

25 mL of an aqueous solution of KCl was found to requires 20 mL of 1M AgNO_(3) solution when titrated using a K_(2)CrO_(4) as indicator. Depression in freezing point of KCl solution with 100% ionisation will be : (K_(f) = 2.0 mol^(-1) kg "and molarity = molality")

What would be the freezing point of aqueous solution containing 17 g of C_(2)H_(5)OH in 100 g of water (K_(f) H_(2)O = 1.86 K mol^(-1)kg) :