Depression in Freezing Point of the Solvent

The freezing point of an aqueous solution of $\mathrm{KCN}$ containing $0.2 \mathrm{mole}/\mathrm{kg}$ water was $-0.80°\mathrm{C}$. On adding $0.1$ mole of $\mathrm{Hg}{\left(\mathrm{CN}\right)}_2$ in the solution containing $1 \mathrm{kg}$ of water, the freezing point of the solution was $-0.6°\mathrm{C}$. Assuming that the complex is formed according to the following equation $\mathrm{Hg}{\left(\mathrm{CN}\right)}_2+{\mathrm{mCN}}^{-}\to \mathrm{Hg}{\left(\mathrm{CN}\right)}_{\mathrm{m}+2}^{\mathrm{m}-}$ and $\mathrm{Hg}{\left(\mathrm{CN}\right)}_2$ is the limiting reactant, the value of $\mathrm{m}$ is

Assertion: Sodium chloride is used to clear snow on the roads.

Reason :Sodium chloride depresses the freezing point of water.

Which of the following statements is false?

The freezing point of an equimolal aqueous solution will be the highest for

If van't Hoff factor for $\mathrm{KCl}$ is $1.8$. What is the boiling point of $0.75 \mathrm{m}$  solution of $\mathrm{KCl}$ in water ${\mathrm{K}}_{\mathrm{b}}=0.5°\mathrm{Ckg}/\mathrm{mol}$?
 

The freezing point $($in ${}^{\circ }\mathrm{C})$ of a solution containing $0.1 \mathrm{g}$ of ${\mathrm{K}}_3\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right] (\mathrm{Mol}.\mathrm{wt}.329)$  in $100 \mathrm{g}$ of water $\left({\mathrm{K}}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\right)$ is :

An aqueous solution containing $6\%$ by weight of urea and $9\%$ by weight of glucose. What will be its approximate freezing point? $\left[{\mathrm{K}}_{\mathrm{f}\left({\mathrm{H}}_2\mathrm{O}\right)}=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\right]$

Which of the following aqueous solution has highest melting point ?

A $0.2$ molal aqueous solution of a weak acid $\left(\mathrm{HX}\right)$ is $20\%$ ionised. The freezing point of this solution is:

The depression in freezing point of $0.01\mathrm{M}$ aqueous ${\mathrm{CH}}_3\mathrm{COOH}$ solution is $0.02046°$. $1 \mathrm{M}$ urea solution freezes at $-1.86°\mathrm{C}$. Assuming molality equal to molarity, $\mathrm{pH}$ of ${\mathrm{CH}}_3\mathrm{COOH}$ solution is :

An aqueous solution of $10\% \mathrm{NaCl}$ (ideal) is cooled. It will allow some

Highest freezing point will be observed for which aqueous solution?

The freezing point of benzene decreases by $0.45°\mathrm{C}$ when $0.2\mathrm{g}$ of acetic acid is added to $20\mathrm{g}$ of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: $\left({\mathrm{K}}_{\mathrm{f}}\right.$for benzene$=5.12 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1})$

The change of energy on freezing $1.00 \mathrm{kg}$ of liquid water of $0°\mathrm{C}$ and $1$ atm is :

$\left.{\left({\mathrm{\Delta H}}_{\text{ice }}\right)}_{\text{fusion }}=6.01 \mathrm{kJ}/\mathrm{mol}\right)$

A $0.001$ molal solution of $\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_4\right]$ in water has a freezing point depression of $0.0054°\mathrm{C}$. If ${\mathrm{K}}_{\mathrm{f}}$ of water is $1.80$, then correct formulation of above molecule is :

A compound is known to be a salt of sodium $\left(\mathrm{NaX}\right)$. If $2 \mathrm{g}$ of salt is dissolved in $100 \mathrm{g}$ of water, the solution freezes at $-1.27°\mathrm{C}$. What will be the compound $\mathrm{NaX}$ ? (${\mathrm{K}}_{\mathrm{f}}$ of water is $1.86 \mathrm{K} \mathrm{kg}/\mathrm{mol}$.)