1.2 mL acetic acid having density 1.06 g cm^(-3) is dissolved in 1 litre of water. The depression in freezing point observed for this concentration of acid was 0.041""^(@)C. The van't Hoff factor of the acid is ( K_(f) of water = 1.86 K kg mol^(-1) )

1.2 mL acetic acid having density 1.06 g cm^(-3) is dissolved in 1 lit

1.	1.2 mL acetic acid having density 1.06 g cm^(-3) is dissolved in 1 litre of water. The depression in freezing point observed for this concentration of acid was 0.041""^(@)C. The van't Hoff factor of the acid is ( K_(f) of water = 1.86 K kg mol^(-1) )
Answer» Solution :Number of moles of acetic acid , `n=(1.2xx1.06)/60=0.0212` Molality of acid solution, `C=0.0212/2=0.0106 "mol kg"^(-1)` `DeltaT_f` expected for this strength = `k_f xx m` =1.86 x 0.016 = 0.0197 K Van.t HOFF FACTOR (i) =`"observed freezing POINT"/"expected freezing point" ="-0.0205"/"-0.0197"`=1.041 Acetic acid is ionised as, `CH_3COOH hArr CH_3COO^(-) + H^(+)` If `.alpha.` is the extent of ionisation, total number of particles =`n(1+alpha)` van.t Hoff factor = i= `(n(1+alpha))/n=1+alpha =1.041, alpha` =0.041 Proton concentration =`C alpha` = 0.0106 x 0.041 `=4.24 xx 10^(-4) "molL"^(-1)`