$2 \ \text{moles}$ each of ethylene glycol and glucose are dissolved in $500 \ \text{g}$ of water. The boiling point of the resulting solution is $:$ (Given $:$ Ebullioscopic constant of water $= 0.52 \ \text{K kg mol}^{-1}$) (in $\text{K}$)

If the mole fraction of the solvent in a solution decreases,then:

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

At $T(K)$,the vapour pressure of pure benzene (molar mass $= 78 \ g \ mol^{-1}$) is $0.85 \ bar$. When $2.0 \ g$ of a non-volatile,non-electrolyte solute is added to $39 \ g$ of benzene,the vapour pressure of the solution at $T(K)$ is $0.83 \ bar$. The elevation in boiling point (in $K$) of the same solution is: ($K_b$ of benzene is $2.6 \ K \ kg \ mol^{-1}$)

Which of the following has the lowest freezing point?

An aqueous solution of a solute $AB$ has a $b.p.$ of $101.08^\circ C$ ($AB$ is $100\%$ ionized at the boiling point of the solution) and freezes at $-1.80^\circ C$. Given $K_b / K_f = 0.3$,the solute $AB$: