In the depression of freezing point experiment,it is found that the:

The ionization constant of a monobasic acid $HA$ is to be determined. If a $0.025 \ m$ aqueous solution of the acid freezes at $-0.060 \ ^{\circ}C$,calculate the ionization constant $K_a$. (Assume molality = molarity and $K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$).

Which of the following graphs correctly represents the changes in thermodynamic properties during the formation of $1 \, mol$ of an ideal binary solution?

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, $K_{b}=0.52 \ K \ kg \ mol^{-1}, \ K_{f}=1.86 \ K \ kg \ mol^{-1}$)

Given below are two statements.
Statement-$I$: Liquids $A$ and $B$ form a non-ideal solution with negative deviation. The interactions between $A$ and $B$ are weaker than $A-A$ and $B-B$ interactions.
Statement-$II$: In reverse osmosis,the applied pressure must be higher than the osmotic pressure of solution.
The correct answer is

For $0.1 \ M$ solutions of urea and $Al_2(SO_4)_3$,which of the following statements is correct? $(a)$ $i$ for urea is $1$ and $i$ for $Al_2(SO_4)_3$ is $5$. $(b)$ $Al_2(SO_4)_3$ has a higher elevation in boiling point. $(c)$ $Al_2(SO_4)_3$ has a higher depression in freezing point. $(d)$ Urea has a higher vapor pressure.