The freezing point of an equimolal aqueous solution will be highest for:

The elevation in the boiling point of an aqueous solution of $NaCl$ is $0.01^{\circ}C$. If its van't Hoff factor is $1.92$,the molality of the $NaCl$ solution is: $(K_{b} \text{ for water} = 0.52 \ K \ kg \ mol^{-1})$ (in $m$)

Which of the following $0.1 \, M$ aqueous solutions has the lowest freezing point?

Calculate the amount of $KCl$ $(M_w = 74.5 \ g \ mol^{-1})$ in grams required to depress the freezing point of $1000 \ g$ of water by $2 \ K$. (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$) (in $.0$)

$A$ $0.1 \ mol$ of a binary electrolyte undergoes complete dissociation in $250 \ g$ of a protic solvent. If the elevation in boiling point is $1.5 \ K$,the molal elevation constant $(K_b)$ is ............. $K \ kg \ mol^{-1}$.

Considering acetic acid dissociates in water,its dissociation constant is $6.25 \times 10^{-5}$. If $5 \ mL$ of acetic acid is dissolved in $1 \ L$ water,the solution will freeze at $-x \times 10^{-2} \ {}^{\circ}C$,provided pure water freezes at $0 \ {}^{\circ}C$.
$x = . . . . . . . . .$ (Nearest integer)
Given: $(K_{f})_{\text{water}} = 1.86 \ K \ kg \ mol^{-1}$.
Density of acetic acid is $1.2 \ g \ mL^{-1}$.
Molar mass of water $= 18 \ g \ mol^{-1}$.
Molar mass of acetic acid $= 60 \ g \ mol^{-1}$.
Density of water $= 1 \ g \ cm^{-3}$.
Acetic acid dissociates as:
$CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$