`0.35` molal aqueous solution of `Na_(x)A` has freezing point of `-3.255^(@)C`. If `K_(f)` of water is `1.86Kkgmol^(-1)`, the value of `x` is: (Assume complete ionisation of salt `Na_(x)A`)

0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If K_(f) for water is 1.86K kg mol^(-1) , the lowering in freezing point of the solution is

A 0.075 molal solution of monobasic acid has a freezing point of -0.18^(@)C . Calculate K_(a) for the acid , (k_(f)=1.86)

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

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If K_(f) for water is 1.86^(@)C//m , the freezing point of the solution will be.

A 0.001 molal aqueous solution of a complex [MA_8] has the freezing point of -0.0054^(@)C . If the primary valency of the salt undergoes 100% ionization and K_f for water =1.8 K molality^(-1) the correct representation of complex is

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):

An aqueous solution of 0.1 molal concentration of sucrose should have freezing point (K_(f)=1.86K mol^(-1)kg)

A 0.001 molal solution of [Pt(NH_(3))_(4)CI_(4)] in water had a freezing point depression of 0.0054^(@)C . If K_(f) for water is 1.80 , the correct formulation for the above molecule is