Colligative Properties Questions

Which of the following aqueous solutions will exhibit the highest boiling point?

The freezing point depression constant for water is $1.86^{\circ }\text{C/m}$. If $5.00\text{g}$ ${\text{Na}}_2{\text{SO}}_4$ is dissolved in $45.0\text{g}$ ${\text{H}}_2\text{O}$, the freezing point is changed by $-3.82^{\circ }\text{C}$. Calculate the van't Hoff factor for ${\text{Na}}_2{\text{SO}}_4$: (Molar mass of ${\text{Na}}_2{\text{SO}}_4=142\text{g/mol}$)

Assertion : The molecular weight of acetic acid determined by depression in freezing point method in benzene and water was found to be different.Reason : Water is polar and benzene is non-polar.

$0.5\text{ molal}$ aqueous solution of a weak acid ($\text{HX}$) is $20\%$ ionised. If $K_f$ for water is $1.86{\text{ K kg mol}}^{-1}$, the lowering in freezing point of the solution is:

A $0.0020\text{m}$ aqueous solution of an ionic compound ${\text{Co(NH}}_3{)}_5({\text{NO}}_2)\text{Cl}$ freezes at $-0.00732^{\circ }\text{C}$. Number of moles of ions which $1\text{mol}$ of ionic compound produces on being dissolved in water will be ($k_f=1.86^{\circ }\text{C/m}$):

A solution of sucrose (molar mass $=342{\text{g mol}}^{-1}$) has been prepared by dissolving $68.5\text{g}$ of sucrose in $1000\text{g}$ of water. The freezing point of the solution obtained will be ($K_f$ for water $=1.86{\text{K kg mol}}^{-1}$):

$0.1\text{molal}$ weak acid is $30\%$ ionized. $K_f=1.86$. Freezing point is:

Of the following $0.10\text{m}$ aqueous solutions, which one will exhibit the largest freezing point depression?

At $100^{\circ }\text{C}$ the vapour pressure of a solution of $6.5\text{g}$ of a solute in $100\text{g}$ water is $732\text{mm}$. If $K_b=0.52$, the boiling point of this solution will be: