The freezing point of a 0.08 molal solution of `NaHSO^(4)` is `-0.372^(@)C`. Calculate the dissociation constant for the reaction.

`K_(f)` for water =`1.86 K m^(-1)`

The freezing point of a 0.08 molal aqueous solutions of NaHSO_(4) is 0.372^(@)C . Calculate the dissociation constant for the reaction HSO_(4)^(-)n hArr H^(+)+SO_(4)^(-2) Assuming no hydrolysis ( k_(t) of water = 1.86 k/kg mole)

The freezing point of a 0.08 molal soluton of NaHSO_4 is -0.372^@C . Calculate the dissociation constant for the reaction: (HSO_4) Leftrightarrow H^+ + SO_4^(2-) (K_f for water= 1.86 Km^-1)

The freezing point of 0.08 molal solution of NaHCO_3 is 272.8 K. Calculate the dissociation constant for the reaction HCO_3^- "iff H^+ + CO_3^(2-) , K_f"(H_2O)=1.86 K mol^-1

The freezing point of a 0.05 molal solution of a non-electrolyte in water is [K_(f)=1.86K//m]

Depression in freezing point of 0.1 molal solution of HF is -0.201^(@)C . Calculate percentage degree of dissociation of HF. (K_(f)=1.86 K kg mol^(-1)) .

The expression in freezing point of a 0.1 molal solution of a substance is 0.372^@ C. what can you say about the solute? (K_f for water is 1.86 K kg mol^(-1))

The freezing points of a 0.05 molal solution of a non-electrolyte in water is: (K, = 1.86 K molality^-1 )