Correct Answer - `0.23k`
`DeltaT = K_(f) xx` molality `xx (1-alpha)`
For acetic acid: `{:(CH_(3)COOH hArr,CH_(3)COOH+,H^(+)),(1,0,0),(1-alpha,alpha,alpha):}`
Given, `alpha = 0.23`, Also molality
`=("mole of acetic acid")/("weight of water in kg")`
`=(3 xx 10^(-3) xx 10^(-3))/(60xx(500 xx 0.997)/(10^(3))) = 0.10`
`DeltaT = K_(f) xx` molality `(1-alpha)`
`DeltaT = 1.86 xx 0.1 xx 1.23 = 0.229`