In the depression of freezing point experiment,it is found that the:

If the mole fraction of the solvent decreases while preparing a solution,then ...........

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 $0 \ ^{\circ}C$)

After adding a non-volatile solute, the freezing point of water decreases to $-0.186^{\circ} C$. Calculate $\Delta T_b$ if $K_f = 1.86 \text{ K kg mol}^{-1}$ and $K_b = 0.521 \text{ K kg mol}^{-1}$. (in $\text{ K}$)

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 .........$^oC$. $[K_f(H_2O) = 1.86 \, K \, kg \, mol^{-1}; K_b(H_2O) = 0.52 \, K \, kg \, mol^{-1}]$

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 $2.6 \ K \ kg \ mol^{-1}$)