0.6 mL of acetic acid (CH_(3)COOH), having density 1.06 g mL^(-1), is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205^(@)C. Calculate the van't Hoff factor and the dissociation constant of acid.

0.6 mL of acetic acid (CH_(3)COOH), having density 1.06 g mL^(-1), is

1.	0.6 mL of acetic acid (CH_(3)COOH), having density 1.06 g mL^(-1), is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205^(@)C. Calculate the van't Hoff factor and the dissociation constant of acid.
Answer» Solution :Number of moles of acetic acid, `=(0.6 mL xx 1.06 g mL^(-1))/(60g mol^(-1))=0.0106 mol = n` Molality `= (0.0106 mol)/(1000 mL xx 1g mL^(-1))` `=0.0106 mol kg^(-1)` `therefore Delta T_(F) = 1.86 " K kg mol"^(-1)xx 0.0106 " mol kg"^(-1)` = 0.0197 K van.t Hoff factor `(i) =("Observed freezing POINT")/("Calculated freezsing point")` `=(0.0205 K)/(0.0197 K)=1.041` Acetic acid is a weak electrolyte and will dissociate into two ions : acetate and hydrogen ions per molecule of acetic acid. If X is the DEGREE of dissociation of acetic acid, then we would have `n(1-X)` moles of undissociated acetic acid, nx moles of `CH_(3)COO^(-)` and nx moles of `H^(+)` ions, `{:(CH_(3)COOH,hArr,H^(+),+,CH_(3)COO^(-)),("n mol",,0,,0),(n(1-x),,"nx mol",,"nx mol"):}` Thus TOTAL moles of particles are: `n(1-x + x + x)=n(1+x)` `i=(n(1+x))/(n)=1+x=1.041` Thus degree of dissociation of acetic acid `=x = 1.041 -1.000=0.041` Then `[CH_(3)COOH]=n(1-x)` `=0.0106(1-0.041)` `[CH_(3)COO^(-)]=nx=0.0106xx0.041`, `[H^(+)]=nx = 0.0106xx0.041` `K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])` `=(0.0106xx0.041xx0.0106xx0.041)/(0.0106(1.00-0.041)=1.86xx10^(-5)`