JEE Main 2019 (Online) 9th January Morning Slot | Solutions Question 112 | Chemistry

The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol^-1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :

37.5 g

75 g

150 g

50 g

JEE Main 2018 (Offline)

MCQ (Single Correct Answer)

-1

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

[Co(H₂O)₃ Cl₃].3H₂O

[Co(H₂O)₆] Cl₃

[Co(H₂O)₅ Cl] Cl₂.H₂O

[Co(H₂O)₄ Cl₂] Cl.2H₂O

JEE Main 2018 (Online) 15th April Evening Slot

MCQ (Single Correct Answer)

-1

Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are M_x and M_y, respectively where M_x = $${3 \over 4}$$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :

$${4 \over 3}$$

$${3 \over 4}$$

$${1 \over 2}$$

$${1 \over 4}$$

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