2g of benzoic acid `(C_(6)H_(5)COOH)` dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62K. Molal depression constant for benzene is 4.9 K kg `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?

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 (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?

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?

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.

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) )

Two elements A and B form compounds having molecular formula AB_(2) and AB_(4) . When dissolved in 20 g of benzene, 1 g of AB_(2) lowers the freezing point by 2.3 K , whereas 1.0 g of AB_(4) lowers it by 1.3 K . The molar depression constant for benzene is 5.1 K kg mol^(-1) . Calculate the atomic mass of A and B .

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:-