Depression of Freezing Point

Choose the correct statement?

De-icing is the process of removing snow, ice or frost from a surface. In extremely cold regions, car windows get covered by ice reducing the visibility. The image below shows the de-icing of the window of a car during extreme cold using a fluid.

Which of the following compounds could be present in the de-icing fluid used above?

An aqueous solution containing $3\mathrm{g}$ of a solute of molar mass $111.6 \mathrm{g} {\mathrm{mol}}^{-1}$ in a certain mass of water freezes at $-0.125°\mathrm{C}$. The mass of water in grams present in solution is $\left({\mathrm{K}}_{\mathrm{f}}=-1.86\mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\right)$

An aqueous solution of sodium chloride freezes below $273 \mathrm{K}$. Explain the lowering in freezing point of water with the help of a suitable diagram.

At $\mathrm{T}\left(\mathrm{K}\right)\mathrm{xg}$ of a non-volatile solid (molar mass $78 \mathrm{g} {\mathrm{mol}}^{-1}$ ) when added to $0.5 \mathrm{kg}$ water, lowered its freezing point by $1.0°\mathrm{C}$. What is $x$ (in $\mathrm{g}$) (${\mathrm{K}}_{\mathrm{f}}$ of water at $\mathrm{T}\left(\mathrm{K}\right)=1.86 {\mathrm{KKgmol}}^{-1}$)

The freezing point depression constant $\left({\mathrm{K}}_{\mathrm{f}}\right)$ of benzene is $5.12\mathrm{Kkg}{ \mathrm{mol}}^{-1}$. Assuming $100\%$ dimerization of acetic acid in benzene, the freezing point depression of $0.2\mathrm{m}$ solution of acetic acid is

$1.02 \mathrm{g}$ of urea when dissolved in $98.5 \mathrm{g}$ of certain solvent decreases its freezing point by $0.211\mathrm{K}$. $1.609 \mathrm{g}$ of an unknown compound when dissolved in $86 \mathrm{g}$ of the same solvent depresses the freezing point by $0.34 \mathrm{K}$. Calculate the molar mass of the unknown compound.

(Molar mass of urea $= 60 \mathrm{g} {\mathrm{mol}}^{-1}$ )

What is a de-icing agent? How does it work?

What freezes out first when a solution of common salt is cooled?

Why is camphor preferred as a solvent in finding the molecular mass of naphthalene by Rast method?

Which of the following solution will have the lowest freezing point and why ? $\left(\mathrm{i}\right) 0.1 \mathrm{M} \mathrm{NaCl}$ solution $\left(\mathrm{ii}\right) 0.1 \mathrm{M}$ glucose solution $\left(\mathrm{iii}\right) 0.1 \mathrm{M} {\mathrm{BaCl}}_2$ solution 

Which of the following solution will have the lowest freezing point and why? $\left(\mathrm{i}\right) 0.1 \mathrm{M} \mathrm{NaCl}$ solution $\left(\mathrm{ii}\right) 0.1 \mathrm{M}$ glucose solution $\left(\mathrm{iii}\right) 0.1 \mathrm{M} {\mathrm{BaCl}}_2$ solution.

What is an antifreeze?

Between $2 \mathrm{M}$ glucose solution and $1 \mathrm{M}$ glucose solution, which one has a lower freezing point?

Give one most important application of the phenomenon of depression in the freezing point in everyday life.

Define molal depression constant/ cryoscopic constant.

$\mathrm{A} \mathrm{solution} \mathrm{of} \mathrm{glucose} (\mathrm{Molar} \mathrm{mass} = 180 \mathrm{g} {\mathrm{mol}}^{-1}) \mathrm{in} \mathrm{water} \mathrm{has} \mathrm{a} \mathrm{boiling} \mathrm{point} \mathrm{of} 100.20°\mathrm{C}. \mathrm{Calculate} \mathrm{the} \mathrm{freezing} \mathrm{point} \mathrm{of} \mathrm{the} \mathrm{same} \mathrm{solution}. \mathrm{Molal} \mathrm{constants} \mathrm{for} \mathrm{water} {\mathrm{K}}_{\mathrm{f}} \mathrm{and} {\mathrm{K}}_{\mathrm{b}} \mathrm{are} 1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1} \mathrm{and} 0.512 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1} \mathrm{respectively}.$

$\mathrm{A} 0.1539 \mathrm{molal} \mathrm{aqueous} \mathrm{solution} \mathrm{of} \mathrm{cane} \mathrm{sugar} (\mathrm{mol}. \mathrm{mass} = 342 \mathrm{g} {\mathrm{mol}}^{-1}) \mathrm{has} \mathrm{a} \mathrm{freezing} \mathrm{point} \mathrm{of} 271 \mathrm{K} \mathrm{while} \mathrm{the} \mathrm{freezing} \mathrm{point} \mathrm{of} \mathrm{pure} \mathrm{water} \mathrm{is} 273.15 \mathrm{K}. \mathrm{What} \mathrm{will} \mathrm{be} \mathrm{the} \mathrm{freezing} \mathrm{point} \mathrm{of} \mathrm{an} \mathrm{aqueous} \mathrm{solution} \mathrm{containing} 5 \mathrm{g} \mathrm{of} \mathrm{glucose} (\mathrm{mol}. \mathrm{mass} = 180 \mathrm{g} {\mathrm{mol}}^{-1}) \mathrm{per} 100 \mathrm{g} \mathrm{of} \mathrm{solution}.$