Question

Pure water freezes at $273K\text{and}1bar$. The addition of $69g$ of ethanol to $500g$ of water changes the freezing point of the solution. Use the freezing point depression constant of water as $2Kkgmo{l}^{-1}$

The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [Molecular weight of ethanol is $46gmo{l}^{-1}$]

Among the following, the option representing change in the freezing point is?

Answer 1

$\Delta T_f=K_f\times m$
where,
$\Delta T_f=T_f^{\circ }-T_f$
$T_f^{\circ }=\text{Freezing point of pure water}=273K$
$T_f=\text{Freezing point of solution}$
$K_f=\text{Freezing Point Depression Constant}=2Kkgmo{l}^{-1}$
$m=\text{Molality= Moles of ethanol per kg of water}$

$\text{Molality}=\frac{69\times 1000}{46\times 500}=3m$

$(T_f^{\circ }-T_f)=K_f\times m\Delta T_f=2\times 3=6K{T}_f=(273-6)K=267K$
Option (a) and (c) have $T_f=267K$
Between (a) and (c):

Also, as the temperature increases, vapour pressure increases, so (a) is incorrect, and (c) is the correct answer