What is the freezing point of a $1 \ molal$ aqueous solution of a non-volatile solute (in $^{\circ} C$)? $(K_{f} = 1.86 \ K \ kg \ mol^{-1}, T_{f}^{\circ} \text{ for water } = 0^{\circ} C)$

What is depression of freezing point? Explain.

The freezing point of a solution containing $1.25 \ g$ of a non-electrolyte solute in $20 \ g$ of water is $271.9 \ K$. What is the molar mass of the solute? (Given: $K_f$ for water = $1.86 \ K \ kg \ mol^{-1}$,Freezing point of pure water = $273 \ K$)

Molal depression constant $(K_{f})$ is dependent on

What is the molar mass of a solute when $5 \ g$ of solute dissolved in $70 \ g$ of solvent lowers its freezing point by $2.5 \ K$? Given $K_f = 3.5 \ K \ kg \ mol^{-1}$.

The freezing point of a $0.05 \ molal$ solution of a non-electrolyte in water is $(K_f = 1.86 \ K \ kg \ mol^{-1})$: