How much amount of $NaCl$ should be added to $600 \ g$ of water $(\rho=1.00 \ g / mL)$ to decrease the freezing point of water to $-0.2^{\circ} C ?$ ............. $gm$
(The freezing point depression constant for water $=2 \ K \ kg \ mol^{-1}$ )

What is the freezing point depression constant of a solvent,$50 \ g$ of which contains $1 \ g$ of a non-volatile solute (molar mass $256 \ g \ mol^{-1}$) and the decrease in freezing point is $0.40 \ K$?

Consider the given plots of vapour pressure $(VP)$ vs temperature $(T / K)$. Which amongst the following options is the correct graphical representation showing $\Delta T_f$,the depression in the freezing point of the solvent in a solution?

Pure water freezes at $273 \ K$ and $1 \ bar$. The addition of $34.5 \ g$ of ethanol to $500 \ g$ of water changes the freezing point of the solution. Use the freezing point depression constant of water as $2 \ K \ kg \ mol^{-1}$. The figures shown below represent plots of vapour pressure $(V.P.)$ versus temperature $(T)$. [Molecular weight of ethanol is $46 \ g \ mol^{-1}$]. Which of the following plots correctly represents the change in the freezing point?

When $x \times 10^{-2} \ mL$ methanol (molar mass $= 32 \ g \ mol^{-1}$; density $= 0.792 \ g \ cm^{-3}$) is added to $100 \ mL$ water (density $= 1 \ g \ cm^{-3}$),the following diagram is obtained.
$x = $ . . . . . . (nearest integer)
[Given: Molal freezing point depression constant of water at $273.15 \ K$ is $1.86 \ K \ kg \ mol^{-1}$]

$2.7 \ kg$ of each of water and acetic acid are mixed. The freezing point of the solution will be $-x^{\circ} C$. Consider the acetic acid does not dimerise in water,nor dissociates in water. $x = . . . . . . .$ (nearest integer)
[Given : Molar mass of water $= 18 \ g \ mol^{-1}$,acetic acid $= 60 \ g \ mol^{-1}$]
$K_f \ H_2O = 1.86 \ K \ kg \ mol^{-1}$
$K_f$ acetic acid $= 3.90 \ K \ kg \ mol^{-1}$
Freezing point: $H_2O = 273 \ K$,acetic acid $= 290 \ K$