Depression in Freezing Point of the Solvent

Assertion: Camphor is usually used in molecular mass determination.

Reason: Camphor has low cryoscopic constant and therefore, causes greater depression in freezing point.

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

Reason :Sodium chloride depresses the freezing point of water.

${\mathrm{PtCl}}_4\mathbf{·}6{\mathrm{H}}_2\mathrm{O}$ can exist as a hydrated complex. $1 \mathrm{molal}$ aqueous solution has a depression in freezing point of $3.72°\mathrm{C}$. Assume, $100\%$ ionization and ${\mathrm{K}}_{\mathrm{f}} \left({\mathrm{H}}_2\mathrm{O}\right)=1.86 \mathrm{K} {\mathrm{mol}}^{-1} \mathrm{kg}$. Then the complex is

When $9.45 \mathrm{g}$ of ${\mathrm{ClCH}}_2\mathrm{COOH}$ is added to $500 \mathrm{mL}$ of water, its freezing point drops by $0.5°\mathrm{C}$. The dissociation constant of ${\mathrm{ClCH}}_2\mathrm{COOH}$ is $\mathrm{x}\times {10}^{-3}$. The value of x is _______. (Rounded off to the nearest integer)

$\left[{\mathrm{K}}_{\mathrm{f}\left({\mathrm{H}}_2\mathrm{O}\right)}=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\right]$

How much amount of $\mathrm{NaCl}$ should be added to $600 \mathrm{g}$ of water $(\mathrm{\rho }=1.00 \mathrm{g}/\mathrm{mL})$ to decrease the freezing point of water to $-0.2°\mathrm{C}?\mathrm{\_\_\_\_\_\_\_}$(The freezing point depression constant for water $=2 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}$)

Round off your answer to the nearest integer after multiplying with 10.

$\mathrm{X} \mathrm{g}$ of NaCl was dissolved in $400 \mathrm{g}$ water and cooled up to $-0.5 °\mathrm{C},$ due to which $28 \mathrm{g}$ of ice is separated out from the solution. Then the value of $\mathrm{x}$ is (Given : ${\mathrm{K}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}$). Assuming complete dissociation of salt. Give answer to the nearest integer.

Find the $\mathrm{pH}$ of $\mathrm{HA}$ aq.solution at $25°\mathrm{C}.$ If its same equimolal aqueous solution compared to urea solution show depression in freezing point in the ratio $12 :10.$$\left(\mathrm{Ka}\right.$ of $\left.\mathrm{HA}={10}^{-5}, \log 2=0.3, \log 3=0.48, \log 5=0.7\right)$

Round off the answer to the nearest integer after multiplying with 10.

Which of the following statements is false?

$31 \mathrm{g}$ of an unknown molecular material is dissolved in $500 \mathrm{g}$ of water. The resulting solution freezes at $271.14 \mathrm{K}$. Calculate molar mass of the material in ${\mathrm{gmol}}^{-1}$.(${\mathrm{K}}_{{\mathrm{f}}_{{\mathrm{H}}_2\mathrm{O}}} = 1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1} , {\mathrm{T}}^{\mathrm{o}}{\mathrm{f}}_{{\mathrm{H}}_2\mathrm{O}}= 273\mathrm{K}$)

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