In a 0.2 molal aqueus solution of a weak acid HX the degree of dissociation is 0.25. The freezing point of the solution will be nearest to: ( K_(f)=1.86 K kg "mol"^(-1) ) a) -0.26^(@)C b) 0.465^(@)C c) -0.48^(@)C d) -0.465^(@)C
0.2 m aqueous solution of a weak acid is 20% dissociated. The boiling point of this solution is (K_(b) of water= 0.52Km^-1)
0.1 M KI and 0.2 M AgNO_(3) are mixed in 3 : 1 volume ratio. The depression of freezing point of the resulting solution will be [ K_(b)(H_(2)O) = 1.86 K kg "mol"^(-1) ]:
An aqueous solution boils at 101^(@)C . What is the freezing point of the same solution? (Gives : K_(f) = 1.86^(@)C// m "and" K_(b) = 0.51^(@)C//m )
Calculate depression of freezing point for 0.56 molal aq. Solution of KCl. (Given : K_f(H_(2)O) = 1.8 kg mol^(-1) ).
PtCl_(4).6H_(2)O can exist as hydrated complex 1 molal aq.solution has depression in freezing point of 3.72^(@)C Assume 100% ionisation and K_(f)(H_(2)O=1.86^(@)mol^(-1))kg then complex is
45g of ethylene glycol (C_2H_6O_2) is mixed with 600 g of water. Calculate: the freezing point of the solution. (K_f for water=1.86K kg mol^-1) .
Out of 1 M urea and 1M KCl solution, which has higher freezing point?
A 0.5 molal solution of ethylene glycol water is used as coolant in a car. If the freezing point constant of water be 1.86^(@)C per mole, the mixture shall freeze at
1.0 g of a monobassic acid HA in 100 g water lowers the freezing point by 0.155 K. IF 0.75 g, of same acid requires 25 mL of N/5 NaOH solution for complete neutralisation then %, degree of ionization of acid is ( K_(f) of H_(2)O = 1.86 K kg "mol"^(-1) ):
NARENDRA AWASTHI-DILUTE SOLUTION-Level 3 - Match The Column
