Two solutions, A and B, are prepared by dissolving a non-volatile solute in different solvents. Solution A has a higher boiling point elevation than solution B, even though both solutions have the same molality. Which of the following statements is the MOST likely explanation for this observation?

Which of the given solutions has highest osmotic pressure?

A solution containing 10 g per ${{\rm{d}}{{\rm{m}}^3}}$ of urea (mol. mass = 60) is isotonic with a 5% (wt.by vol.) solution of a non-volatile solute. The molecular mass (in ${\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$) of non-volatile solute is :

On adding 1 g arsenic to 80 g benzene, the freezing point of benzene is lowered by $0.19^\circ C$ . The formula of arsenic is

A 0.5 molal aqueous solution of a weak acid (HX) is 20 per cent ionized. The lowering in freezing point of this solution is :

$\left( {{K_f} = 1.86\;{\rm{K}}/{\rm{m}}\;{\rm{ for }}\;{\rm{water}}} \right)$

0.1 molal aqueous solution of NaBr freezes at $- 0.335^\circ C$ at atmospheric pressure ${k_f}$ for water is $1.86^\circ C$. The percentage of dissociation of the salt in solution is

The empirical formula of a non-electrolyte is $C{H_2}O$. A solution containing 6g of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution at the same temperature. The molecular formula of the compound is

A solution of 4.5 g of a pure non-electrolyte in 100 g of water was found to freeze at $0.465^\circ C$. The molecular weight of the solute closest to $({k_f} = 1.86)$

Three solutions were prepared by dissolving 5g of glucose ($C_6H_{12}O_6$) in 125 ml of water ($P_1$), 5g of Urea ($CH_4N_2O$) in 125 ml of water ($P_2$), and 5g of sucrose ($C_{12}H_{22}O_{11}$) in 125 ml of water ($P_3$). Arrange the solutions in decreasing order of osmotic pressure.

At a constant temperature, which of the following aqueous solutions will have the maximum vapour pressure?

(Mol. wt NaCl =58.5, ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} = 98.0\;{\rm{g}}{\rm{.mo}}{{\rm{l}}^{ - 1}})$

Vapour pressure of chloroform $\left( {{\rm{CHC}}{{\rm{l}}_3}} \right)$ and dichloromethane $\left( {{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{l}}_2}} \right)\;\;at\;25^\circ C$ are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of $\left( {{\rm{CHC}}{{\rm{l}}_3}} \right)$ and 40 g of ${{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{l}}_2}}$ at the same temperature will be :

(Molecular mass of ${\rm{CHC}}{{\rm{l}}_3} = 119.5\;\;u$ and molecular mass of ${\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{l}}_2} = 85\;u)$