1.2 ~mL of 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 )

Step-by-step solution

From depression in freezing point, aligned & Molality (m)= Moles of solute Mass of solvent ( in g ) 1000 & Mass = density volume =1.2 1.06=1.272 ~g & Moles of solute = 1.272 60 & ( Moles = Mass Molecular mass ) & 0.041= i 1.86 1.272> 1000 60 1000 = 60 0.041 1.86 1.27 & i=1.04 aligned Thus, option (b) is correct.