The freezing point depression of 0.1 molal NaCl solution is 0.372 K. What conclusion can you draw about the molecular state of NaCl in water. `K_f` of water = 1.86 k/m.

The boiling point elevation of 600 mg of acetic acid in 0.1 kg of benzene is 0.1265 K. What conclusion can you draw about the molecular state of acetic acid in benzene ? K_b of benzene is 2.53 k//m.

The freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is:

The freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is:

The freezing point depression of 0.1 molal solution of acetic acid in benzene is 0.256 K , K_(f) for benzene is 5.12 K Kg mol^(-1) . What conclusion can you draw about the molecular state of acetic acid in benzence ?

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)

Calculate the freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water. ( K_(f) for water = 1.86^(@)"mol"^(-) ).

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

Calculate the freezing point depression and boiling point elevation of a solution of 10.0 g of urea (M_(B)=60) in 50.0 g of water at 1 atm . pressure. K_(b) and K_(f) for water 0.52^(@)C m^(-1) and 1.86^(@)C m^(-1) respectively.

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

If the freezing point of 0.1 molal solution of X in water is more than 0.1 molal solution of Y in water then which among the following conclusion can be drawn?