If the freezing point of `0.1 M HA(aq)` solution is `-0.2046^(@)C` then `pH` of solution is
`(` If `K_(f)` water `=1.86mol^(-1)kg^(-1))`

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)) .

For an aqueous solution freezing point is -0.186^(@)C . The boiling point of the same solution is (K_(f) = 1.86^(@)mol^(-1)kg) and (K_(b) = 0.512 mol^(-1) kg)

For an aqueous solution freezing point is -0.186^(@)C . The boiling point of the same solution is (K_(f) = 1.86^(@)mol^(-1)kg) and (K_(b) = 0.512 mol^(-1) kg)

For an aqueous solution freezing point is -0.186^(@)C . The boiling point of the same solution is (K_(f) = 1.86^(@)mol^(-1)kg) and (K_(b) = 0.512 mol^(-1) kg)

The boiling point of an aqueous solution is 100.18 .^(@)C . Find the freezing point of the solution . (Given : K_(b) = 0.52 K kg mol^(-1) ,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 point of 0.20M solution of weak acid HA is 272.5K . The molality of the solution is 0.263mol Kg^(-1) . Find the pH of the solution on adding 0.20M solution of NaOH . Given: K_(f) of water = 1.86 Km^(-1)

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):