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)` )

Options

A. 0.41
B. 1.04
C. 0.96
D. 1.54

Answer

B

Explanation

From depression in freezing point,
` DeltaT_(f) = I xx K_(f) xx m " " `…. (i)
Molality (m) = `("Moles of solute")/("Mass of solvent (in g)") xx 1000 `
Mass = density `xx` volume = 1.2 `xx 1.06 = 1.272 ` g
`therefore` Moles of solute = `(1.272)/(60)`
`(because " moles" = ("mass")/("molecular mass")) `
`0.041 = (i xx 86 xx 1.272 gt xx 1000)/(60 xx 1000) = (60 xx 0.041)/(1.86 xx 1.27)`
i = 1.04
Thus, option (b) is correct.