`2g` of benzoic acid `(C_(6)H_(5)COOH)` dissolved in `25g` of benzene shows a depression in freezing point equal to `1.62K`.Molal depression constant for benzene is `4.9Kkgmol^(-1)`.What is the percentage association of acid if it forms dimer in solution?

Two grams of benzoic acid (C_(6)H_(5)COOH) dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K . Molal depression constant for benzene is 4.9 K kg^(-1)"mol^-1 . What is the percentage association of acid if it forms dimer in solution?

Two grams of benzoic acid (C_(6)H_(5)COOH) dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K . Molal depression constant for benzene is 4.9 K kg^(-1)"mol^-1 . What is the percentage association of acid if it forms dimer in solution?

2g of benzoic acid dissolved in 25g of benzene shows a depression in freezing point equal to 1.62K. What is the percentage association of benzoic acid if it forms a dimer in solution ? ( K_(f) for benzene = 4.9 K kg "mol"^(-1) )

Molecules of benzoic acid (C_(6)H_(5)COOH) dimerise in benzene. ‘w’ g of the acid dissolved in 30g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimer in the solution is 80, then w is : (Given that K_(f)=5K Kg "mol"^(-1) Molar mass of benzoic acid =122 g "mol"^(-1) )

1.0 g of non-electrolyte solute dissolved in 50.0 g of benzene lowered the freezing point of benzene by 0.40 K . The freezing point depression constant of benzene is 5.12 kg mol^(-1) . Find the molecular mass of the solute.

Calculate molality of 2.5g of ethanoic acid (CH_(3)COOH) in 75 g of benzene.

2.44 g 'w' g of the benzoic acid dissolved in 30 g of benzene shows depression in freezing equal to 2K. If the percentage association of the acid to form polymer (C_(6)H_(5)COOH)_(n) in the solution is 80, then find numerical value of n. (Given that K_(f) = 5K kg mol^(-1) , Molar mass of benzoic acid = 122 g mol^(-1) )

1.00 g of a non-electrolyte solute (molar mass 250g mol^(–1) ) was dissolved in 51.2 g of benzene. If the freezing point depression constant K_(f) of benzene is 5.12 K kg mol^(–1) , the freezing point of benzene will be lowered by:-

1.00 g of a non-electrolyte solute (molar mass 250g mol^(–1) ) was dissolved in 51.2 g of benzene. If the freezing point depression constant K_(f) of benzene is 5.12 K kg mol^(–1) , the freezing point of benzene will be lowered by:-

When 20 g of napthanoic acid (C_(11)H_(8)O_(2)) is dissolved in 50 g of benzene (K_(f)=1.72 K kg/mol) a freezing point depression of 2 K is observed. The van't Hoff factor (i) is