Distilled water boils at $373.15 \ K$ and freezes at $273.15 \ K$. $A$ solution of glucose in distilled water boils at $373.202 \ K$. What is the freezing point (in $K$) of the same solution? (For water, $K_{b}=0.52 \ K \ kg \ mol^{-1}, \ K_{f}=1.86 \ K \ kg \ mol^{-1}$)

The difference between the boiling point and freezing point of an aqueous solution of urea containing $10.0 \ kg$ of water is $100.2372 \ ^oC$. How many grams of urea are dissolved in the solution? (Given: $K_b = 0.513 \ K \ kg \ mol^{-1}$,$K_f = 1.86 \ K \ kg \ mol^{-1}$)

Properties such as boiling point,freezing point,and vapour pressure of a pure solvent change when solute molecules are added to get a homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of an ethylene glycol and water mixture as an anti-freezing liquid in the radiator of automobiles.
$A$ solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is $0.9$.
Given: Freezing point depression constant of water $(K_{f}^{\text{water}}) = 1.86 \ K \ kg \ mol^{-1}$
Freezing point depression constant of ethanol $(K_{f}^{\text{ethanol}}) = 2.0 \ K \ kg \ mol^{-1}$
Boiling point elevation constant of water $(K_{b}^{\text{water}}) = 0.52 \ K \ kg \ mol^{-1}$
Boiling point elevation constant of ethanol $(K_{b}^{\text{ethanol}}) = 1.2 \ K \ kg \ mol^{-1}$
Standard freezing point of water $= 273 \ K$
Standard freezing point of ethanol $= 155.7 \ K$
Standard boiling point of water $= 373 \ K$
Standard boiling point of ethanol $= 351.5 \ K$
Vapour pressure of pure water $= 32.8 \ mm \ Hg$
Vapour pressure of pure ethanol $= 40 \ mm \ Hg$
Molecular weight of water $= 18 \ g \ mol^{-1}$
Molecular weight of ethanol $= 46 \ g \ mol^{-1}$
In answering the following questions,consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.
$1.$ The freezing point of the solution $M$ is
$(A) \ 268.7 \ K \ (B) \ 268.5 \ K$
$(C) \ 234.2 \ K \ (D) \ 150.9 \ K$
$2.$ The vapour pressure of the solution $M$ is
$(A) \ 39.3 \ mm \ Hg \ (B) \ 36.0 \ mm \ Hg$
$(C) \ 29.5 \ mm \ Hg \ (D) \ 28.8 \ mm \ Hg$
$3.$ Water is added to the solution $M$ such that the mole fraction of water in the solution becomes $0.9$. The boiling point of this solution is
$(A) \ 380.4 \ K \ (B) \ 376.2 \ K$
$(C) \ 375.5 \ K \ (D) \ 354.7 \ K$
Give the answer for questions $1, 2$ and $3.$

For a solution formed by mixing liquids $L$ and $M$,the vapour pressure of $L$ plotted against the mole fraction of $M$ in solution is shown in the following figure. Here $x_L$ and $x_M$ represent mole fractions of $L$ and $M$,respectively,in the solution. The correct statement$(s)$ applicable to this system is(are)
$A$. Attractive intermolecular interactions between $L-L$ in pure liquid $L$ and $M-M$ in pure liquid $M$ are stronger than those between $L-M$ when mixed in solution
$B$. The point $Z$ represents vapour pressure of pure liquid $M$ and Raoult's law is obeyed when $x_L \rightarrow 0$
$C$. The point $Z$ represents vapour pressure of pure liquid $L$ and Raoult's law is obeyed when $x_L \rightarrow 1$
$D$. The point $Z$ represents vapour pressure of pure liquid $M$ and Raoult's law is obeyed from $x_L=0$ to $x_L=1$

$20\%$ of acetic acid is dissociated when its $5\ g$ is added to $500\ mL$ of water. The depression in freezing point of such water is $..... \times 10^{-3} \ { }^{\circ}C$. Atomic mass of $C$,$H$,and $O$ are $12$,$1$,and $16 \ \text{a.m.u.}$ respectively.
[Given: Molal depression constant and density of water are $1.86 \ K \ kg \ mol^{-1}$ and $1 \ g \ cm^{-3}$ respectively.]

Assertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.
Reason : Raoult's law is a special case of Henry's law.