The freezing point of a solution prepared from $1.25 \ g$ of a non-electrolyte and $20 \ g$ of water is $271.9 \ K$. If the molar depression constant is $1.86 \ K \ kg \ mol^{-1}$,then the molar mass of the solute will be:

$150 \ g$ of acetic acid was contaminated with $10.2 \ g$ ascorbic acid $(C_{6}H_{8}O_{6})$ to lower its freezing point by $(x \times 10^{-1})^{\circ} C$. The value of $x$ is (Nearest integer). [Given $K_{f} = 3.9 \ K \ kg \ mol^{-1}$; Molar mass of ascorbic acid $= 176 \ g \ mol^{-1}$]

$1.00 \ g$ 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 ............ $g \ mol^{-1}$.

If molality of the dilute solution is doubled,the value of molal depression constant $(K_f)$ will be

Which of the following aqueous solutions has the highest freezing point?

How many grams of methyl alcohol should be added to a $10 \, L$ tank of water to prevent its freezing at $268 \, K$? ($K_f$ for water is $1.86 \, K \, kg \, mol^{-1}$,density of water is $1 \, kg/L$)