An aqueous solution freezes at $-0.186 \ ^\circ C$ ($K_f = 1.86 \ ^\circ C \ kg \ mol^{-1}$; $K_b = 0.512 \ ^\circ C \ kg \ mol^{-1}$). What is the elevation in boiling point of the same solution?

After adding a solute,the freezing point of the solution decreases to $-0.186 \ ^{\circ}C$. Calculate $\Delta T_b$ if $K_f = 1.86 \ K \ kg \ mol^{-1}$ and $K_b = 0.521 \ K \ kg \ mol^{-1}$. (Assume the freezing point of pure solvent is $0 \ ^{\circ}C$)

An aqueous solution of $NaCl$ shows the depression of freezing point of water equal to $0.372 \, K$. The boiling point of $BaCl_2$ solution of same molality will be .........$^oC$. $[K_f(H_2O) = 1.86 \, K \, kg \, mol^{-1}; K_b(H_2O) = 0.52 \, K \, kg \, mol^{-1}]$

$1.22 \, g$ of an organic acid is separately dissolved in $100 \, g$ of benzene $(K_{b}=2.6 \, K \, kg \, mol^{-1})$ and $100 \, g$ of acetone $(K_{b}=1.7 \, K \, kg \, mol^{-1})$. The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by $0.17^{\circ} C$.
The increase in boiling point of solution in benzene in $^{\circ} C$ is $x \times 10^{-2}$. The value of $x$ is ..... .(Nearest integer)
$[$ Atomic mass : $C=12.0, H=1.0, O=16.0]$

$W \ g$ of a non-volatile electrolyte solid solute of molar mass $M \ g \ mol^{-1}$ when dissolved in $100 \ mL$ water, decreases vapor pressure of water from $640 \ mm \ Hg$ to $600 \ mm \ Hg$. If aqueous solution of the electrolyte boils at $375 \ K$ and $K_b$ for water is $0.52 \ K \ kg \ mol^{-1}$, then the mole fraction of the electrolyte solute $(X_2)$ in the solution can be expressed as (Given density of water $= 1 \ g/mL$ and boiling point of water $= 373 \ K$):

$A$ solution of urea in water has a boiling point of $100.18^{\circ} C$. What is the freezing point of the same solution,if $K_{f}$ and $K_{b}$ of water are $1.86$ and $0.52 \ K \ kg \ mol^{-1}$,respectively (in $^{\circ} C$)? (Boiling point of water $= 100^{\circ} C$ )