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?

A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g of sucrose per litre. The molecular mass of protein is :

A solution containing ${\rm{ }}4\;g$ of polyvinyl chloride polymer in one litre of dioxane was found to have an osmotic pressure of $4.1 imes {10^{ - 4}}\;atm$ at $27^\circ C$ . The approximate molecular weight of the polymer is

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

The molecular weight of NaCl determined by studying freezing point depression of its 0.5% aqueous solution is 30. The apparent degree of dissociation of NaCl is

At $25^\circ C$ , the highest osmotic pressure is exhibited by $0.1\;M$ solution of

Calculate the molal depression constant of a solvent which has freezing point $16.6^\circ C$ and latent heat of fusion$180.75\;J\;{g^{ - 1}}$.

The amount of ice that will separate out on cooling a solution containing 50 g of ethylene glycol in 200 g water to $- 9.3^\circ {\rm{C}}$ is : ({{K'}_f} = 1.86\;{\rm{K}}\;{\rm{molalit}}{{\rm{y}}^{ - 1}})

In a 0.2 molal aqueous solution of a weak acid HX, the degree of ionisation is 0.3 Taking ${k_f}$ for water as 1.85, the freezing point of the solution will be nearest to

Abnormal colligative properties are observed only when the dissolved non-volatile solute in a given dilute solution

At 298 K, the plot of osmotic pressure ($\Pi$) vs. concentration ($\mathrm{mol\ L^{-1}}$) for a solution gives a straight line. If the slope is $24.7 \mathrm{L\ bar\ mol^{-1}}$, what is the value of the ideal gas constant (R) being used?