The freezing point of a `0.50` molal aq.solution of weak acid (HA) is `-1.302^(@)C`.The dissociation constant of acid HA is [Given: `K_(f)` for water `=1.86kgmol^(-1)K]`

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 0.05 m solution of glucose in water is (K1 = 1.86°C m^(-1) )

The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91^(@)C . The freezing point constant of water, K_(f) , is 1.86 K kg mol^(-1) . The percentage dissociation of HF at this concentration is:-

If 0.2 molal aqueous solution of a weak acid (HA) is 40% ionised then the freezing point of the solution will be (K_f for water = 1.86degC/m

% dissociation of a 0.024M solution of a weak acid HA(K_(a)=2xx10^(-3)) is :

The freezing point of a 0.05 molal solution of a non-electrolyte in water is: ( K_(f) = 1.86 "molality"^(-1) )

0.01 m aqueous solution of K_(3)[Fe(CN)_(6)] freezes at -0.062^(@)C . What is the apparent percentage of dissociation ? ( K_(f) for water = 1.86" K kg mol"^(-1) )

A 0.2 molal aqueous solution of a weak acid HX is 20% ionized. The freezing point of the solution is (k_(f) = 1.86 K kg "mole"^(-1) for water):

A 0.2 molal aqueous solution of a weak acid HX is 20% ionized. The freezing point of the solution is (k_(f) = 1.86 K kg "mole"^(-1) for water):

Determine the freezing point (nearest integer) of a 1 mole lkg^(-1) aqueous solution of a weak electrolyte that is 7.5% dissociated into two ions (in .^(@)C ) [Given K_(f) of water is 1.86^(@)C//m ].