QJEEAdv08Ch8bf9: JEE Advanced 2008 Chemistry - Class 12 JEEAdv08Ch8bf9 Solution: The freezing point of the solution M is :

For solution M, water is solute and ethanol is solvent.

$\because$ $\Delta {T_f} = {({K_f})_{ethanol}} \times m$

$2 \times {{0.1} \over {0.9 \times {{46} \over {1000}}}} = {{2000} \over {9 \times 46}} = 4.8$

$\therefore$ ${T_f} = T_f^0 - \Delta {T_f} = 155.7 - 4.8 = 150.9\,K$