JEE (Advanced) 2008 Paper-1

In the following reaction sequence, products I, J and L are formed. K represents a reagent.

There are some deposits of nitrates and phosphates in earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridization easily explains the ease of sigma donation capability of $$NH_3$$ and $$PH_3$$. Phosphine is a flammable gas and is prepared from white phosphorous.

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examplesis the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles
A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtureis 0.9
Given : Freezing point depression constant of water $$\left(K_{f}^{water}\right)$$ = 1.86 K kg mol$$^{-1}$$
Freezing point depression constant of ethanol $$\left(K_{f}^{ethanol}\right)$$ = 2.0 K kg mol$$^{-1}$$
Freezing point elevation constant of water $$\left(K_{b}^{water}\right)$$ = 0.52 K kg mol$$^{-1}$$
Freezing point elevation constant of ethanol $$\left(K_{b}^{ethanol}\right)$$ = 1.2 K kg mol$$^{-1}$$
Standard freezing point of water = 273 K
Standard freezing point of ethanol = 155.7 K
Standard boiling point of water = 373 K
Standard boiling point of ethanol = 351.5 K
Vapour pressure of pure water = 32.8 mm Hg
Vapour pressure of pure ethanol = 40 mm Hg
Molecular weight of water = 18 g mol$$^{-1}$$
Molecular weight of ethanol = 46 g mol$$^{-1}$$

In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.