Calculate elevation in boiling point for 2 molal aqueous solution of glucose. (Given K_b(H_(2)O) = 0.5 kg mol^(-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) ]:
Calculate the freezing point of a one molar aqueous solutoin (density 1.04gL^-1) of KCl. ( K_f of water=1.86Kg mol^-1 , atomic masses of K=39, Cl=35.5)
if the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associanting solute in solvent ( K_(b) =x K kg "mol"^(-1) )is yK,then the depression in freezing point of solution of same concentration would be ( K_(f) )of the sovent = zk.kg"mol"^(-1) )
In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):
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
True or false The depression in freezing point for 1 m solution of a solute in water and benzene is same.
2g of benzoic acid ( C_6H_5COOH ) is dissolved in 25g of benzene show depression in freezing point equal to 1.62K. Molar depression constant for benzene, K_f =4.9K kg mol^-1 . What is percentage association of acid if it forms a dimer in solution?
The freezing point of a solution containing 0.1g of K_3[Fe(CN)_6] in 100 g of water. (K_f=1.86 K kg mol^-1) is
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
