When a solute is dissolved in a solvent,the freezing point decreases by $0.184 \ ^oC$. What will be the molality of the solution? (Given $K_f = 18.4 \ K \ kg \ mol^{-1}$)

If the freezing point of a $5\%$ (by mass) aqueous solution of cane sugar is $271 \ K$ and the freezing point of pure water is $273.15 \ K$,then the freezing point of a $5\%$ (by mass) aqueous solution of glucose will be .......... $K$.

Calculate the molality of a solution having freezing point depression $3.6 \ K$ and freezing point depression constant $4.8 \ K \ kg \ mol^{-1}$.

$A$ camphor sample melts at $176^{\circ} C$. $K_f$ for camphor is $40 \ K \ kg \ mol^{-1}$. $A$ solution of $0.02 \ g$ of a hydrocarbon in $0.8 \ g$ of camphor melts at $156.77^{\circ} C$. The hydrocarbon is made up of $92.3 \%$ of carbon. What is the molecular formula of the hydrocarbon?

When $1.25 \ g$ of a non-volatile solute is dissolved in $20 \ g$ of water,the freezing point of the solution is found to be $271.9 \ K$. If the molal depression constant $(K_f)$ is $1.86 \ K \ kg \ mol^{-1}$,what is the molar mass of the solute?

The freezing point depression of $645 \ g$ of an aqueous solution of ethylene glycol $(C_2H_6O_2)$ is $2.25 \ K$. Find the weight of ethylene glycol in the solution. $[K_f = 1.86 \ K \ kg \ mol^{-1} ; H = 1, C = 12, O = 16 \ amu]$ (in $g$)