NEET Chemistry MCQs

For an aqueous solution, freezing point is $-0.186^\circ$. Elevation of the boiling point of the same solution is $({k_f} = 1.86^\circ \;{\rm{ }}mo{l^{ - 1}}\;and\;{k_b} = 0.512^\circ \;{\rm{ }}mo{l^{ - 1}}\;kg)$

The vapour pressure of benzene at $90^\circ C$ is 1020 torr. A solution of 5 g of a solute in 58.5 g benzene has vapour pressure 990 torr. The molecular weight of the solute is :

1.0 g of a non-electrolyte solute (molar mass 250 g ${\rm{mo}}{{\rm{l}}^{ - 1}}$) was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is 5.12 K kg ${\rm{mo}}{{\rm{l}}^{ - 1}}$, the lowering in freezing point will be :

Which one of the statements given below concerning properties of solutions, describes a colligative effect?

1.0 g of a non-electrolyte solute (molar mass 250 g ${\rm{mo}}{{\rm{l}}^{ - 1}}$) was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is 5.12 K kg ${\rm{mo}}{{\rm{l}}^{ - 1}}$, the lowering in freezing point will be :

Which of the given solutions has highest osmotic pressure?

The molal elevation constant for water is ${\rm{0}}{\rm{.52}}\;{\rm{K}}\;{\rm{molalit}}{{\rm{y}}^{ - 1}}$. The elevation caused in the boiling point of water by dissolving 0.25 mole of a non-volatile solute in 250 g of water will be :

Which solution will have least vapour pressure?

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

Which of the given solutions has highest osmotic pressure?