The freezing point of solution containing `0.2 g` of acetic acid in `20.0 g` of benzene is lower...

The freezing point of a solution containing of 0.2g of acetic acid in 20.0g of benzene is lowered 0.45^(@)C . Calculate. (i) the molar mass of acetic acid from this data (ii) Van't Hoff factor [For benzene, K_(f)=5.12K kg "mol"^(-1) ] What conclusion can you draw from the value of Van't Hoff factor obtained ?

The freezing point of a solution contaning 0.3 g of acetic acid in 43 g of benzene reduces by 0.3^(@) . Calculate the Van's Hoff factor "( K_(f) for benzene = 5.12 K kg mol^(-1) )"

The freezing point of a solution containing 0.3 g of acetic acid in 30 g of acetic acid in 30 g of benzene is lowerd by 0.54^(@) , Calculate Van't Hoff factor. (K_(f)" for benzene"= 5.12 " K kg mol"^(-1) ).

The freezing point of a solution containing 2.40 g of a compound in 60.0g of benzene is 0.10^(@)"C" lower than that of pure benzene. What is the molecular weight of the compound? ("K"_("f")" is " 5.12^(@)"C"//"m for benzene"

Phenol associates in benzene to from dimer (C_(6)H_(5)OH)_(2) . The freezing point of a solution containing 5g of phenol in 250 g of benzene is lowered by 0.70^(@)C . Calculate the degree of association of phenol in benzene .( k_(f) for benzene = 5.12 Km^(-1) )

Calculate the molecular weight of a substance if the freezing point of a solution containing 100 g of benzene and 0.2 g of the substance is 0.17 K below that of benzene. The cryoscope constant of benzerne is 5.16 K kgmol^(-1)