`0.6 mL` of acetic acid `(CH_(3)COOH)` having density `1.06 g mL^(-1)` is dissolved in `1 L` of water. The depression in freezing point observed for this strength of acid was `0.0205^(@)C`.Calculate the Van't Hoff factor and dissociation constant of the acid. `K_(f)` for `H_(2)O=1.86 K kg ^(-1) "mol"^(-1))`

Explanation

Moles of acetic acid ` = (0.6 mL xx 1.06 g mL)/(60 g mol)`
` = 0.016 mol`
`[M(CH_(3)COOH = 60)]`
Mass of water ` = 1000 mL xx 1 g mL^(-1) - 1000g `
Molality ` = (0.0106 )/(1000g) xx 1000 = 0.0106 mol kg^(-1)`
Now `Delta T_(f) = k_(f) xx m `
` = 1.86 K kg mol^(-1) xx 0.0106 mol Kg^(*-1)`
` = 0.0197 K`
Observed freezing point depression = 0.0205 K .
Van.t Hoff factor `= ("Observed freezing point ")/("Calculated freezing point")`
` = (0.0205 K)/(0.0197 K) = 1.041`
Acetic acid dissociates as :
`CH_(3)COOH hArr CH_(3)COO^(-) + H^(+)`
`{:(,"Initial conc.",m,0,0),(,"Conc.after","m(1-xmol)", "mx mol" , "mx mol"):}`
dissociation .
Total moles of particle ` = m (1-x+x + x) = m(1+x)`
`i= (m(1+x))/(m) = 1.041`
` therefore " " x = 0.041`
`therefore ` Degree of dissociation of = 0.041
`[CH_(3)COOH] = m (1-x) = 0.0106 xx (1-0.041) = 0.0102 `
`[CH_(3)COO^(-)] = mx = 0.0106 xx 0.041 = 4.35 xx 10^(-4)`
`[H^(+)] = mx = 4.35 xx 10^(-4)`
` K_(b) = ([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])`
` = ((4.35 xx 10^(-4)) (4.35 xx 10^(-4)))/(0.0102)`
` = 1.86 xx 10^(-5)`