Calculate the molality of a solution if its depression in freezing point is $0.18 \ K$. $[K_f = 1.6 \ K \ kg \ mol^{-1}]$ (in $m$)

The molal depression constant for water is $1.86 \, ^\circ C \, kg \, mol^{-1}$. The freezing point of a $0.05 \, m$ solution of a non-electrolyte in water is:

If $6 \ g$ of solute dissolved in $100 \ g$ of water lowers the freezing point by $0.93 \ K$. What is the molar mass of the solute? $(K_{f} = 1.86 \ K \ kg \ mol^{-1})$

What is depression of freezing point? Explain.

$1.00 \ g$ of a non-electrolyte solute dissolved in $50 \ g$ of benzene lowered the freezing point of benzene by $0.40 \ K$. The ${K_f}$ for benzene is $5.12 \ K \ kg \ mol^{-1}$. The molecular mass of the solute will be ............ $g \ mol^{-1}$.

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 $K_f = 3.5 \ K \ kg \ mol^{-1}$.