The cryoscopic constant of water is `1.86`. The elevation of b.p. for a `0.1m` solution of an unknown solute in water is

The molal depression constant for water is 1.86^(@)C . The freezing point of a 0.05-molal solution of a non-electrolyte in water is

0.1 M solution of a solute has a water potential of

The cryoscopic constant of water is 1.86 K "mol"^(-1) kg. An aqueous solution of cane sugar freezes at -0.372^(@)C . Calculate the molality of the solution.

The cryoscopic contant of water is 1.86 K kg mol^(-1) . A 0.01 molal acetic acid solution produces a depression of 0.0194^(@)C in the freezing point. The degree of dissociation of acetic acid is :

Calculate the mole fraction of an unknown solute in 2.0 m aqueous solution.

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

If K_f "value of " H_2 O is 1.86 ^@C /molal. The value of DeltaT_f for 0.1 m solution of non-volatile solute is

0.1 molal solution of Hg(NO_(3))_(2) freezes at 0.558^(@) C. The cryoscopic constant for water is 1.86 KKg mol^(-1) then what will be the percentage ionisation of salt ?

The molal elevation constant of water = 0.52 K m^(-1) . The boiling point of 1.0 molal aqueous KCl solution (assuming complete dissociation of KCl ) should be

The molal elevation constant of water = 0.52 K m^(-1) . The boiling point of 1.0 molal aqueous KCl solution (assuming complete dissociation of KCl ) should be