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?

Water will boil at 101.5$^\circ C$ at which of the following pressure?

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

A solution containing 1 g of a non-volatile solute in 100 g of camphor lowers the freezing point of camphor by 0.4 °C. If the molal depression constant ($K_f$) for camphor is 40 K kg/mol, what is the molar mass of the solute?

The freezing point of one modal NaCl solution assuming NaCl to be 100 % dissociated in water is (modal depression constant=1.86)

The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $\Delta {T_f}$, when 0.01 mole of sodium sulphate is dissolved in 1 kg of water, is $({k_f} = 1.86\;{\rm{K}}\;{\rm{kg mo}}{{\rm{l}}^{ - 1}})$

Depression in freezing point is 6 K for NaCl solution if ${k_f}$ for water is 1.86 K/kg mol, amount of NaCl dissolved in 1 kg water is

Y g of non-volatile organic substance of molecular mass M is dissolved in 250 g benzene. Molal elevation constant of benzene of ${{K_b}}$. Elevation in its boiling point is given by :

The osmotic pressure of a solution at $0^\circ C$ is 2 atm. What will be its osmotic pressure at $273^\circ C$ under similar conditions?

The osmotic pressure $(At\;27^\circ C)$ of an aqueous solution (200 mL) containing 6 g of a protein is $2 imes {10^{ - 3}}\;{\rm{atm}}.$ If $R = 0.080{\rm{ }}L - atm{\rm{ }}\;mo{l^{ - 1}}{K^{ - 1}},$ the molecular weight of protein is