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Give reasons for the following: (a) At h...

Give reasons for the following: (a) At higher altitudes, people suffer from a disease called anoxia. In this disease, they become weak and cannot think clearly. (b) When mercuric iodide is added to an aqueous solution of KI, the freezing point is raised.

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(a) At higher altitudes, available oxygen is less, causing anoxia.
(b) ` underset("1mol")(HgI_2) + underset("2 mol")(2KI) to underset("potassium tetraiodomercurate (II)")(K_2HgI_4)`
Due to the formation of complex compound `K_2HgI_4` , the number of moles of the solute in the solution decreases. Greater the number of moles of the solute in the solution, greater the depression in freezing point. Smaller the number, the reverse effect will take place
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Addition of non-volatile solute to solvent lowers its vapoure pressure. Therefore, the vapour pressure of a solution (i.e, V.P. of solvent in a solution) is lower than that of pure solvent in a solution) is lower than that of pure solvent, at the same temperature. A higher temperature is needed to raise the vapour pressure upto one atmosphere pressure, when boiling point is attined. However, increase in b.pt. is small . for example, 0.1 molal aqueous sucrose solution boils at 10.05^(@)C Sea water, an aqueous solution, which is rich in Na^(+) and Cl^(-) ions, freezes about 1^(@)C lower than frozen water . At the freezing point of a pure-solvent, the reates at which two molecule stick together to form the solid and leave it to return to liquid are equal when solute is present. Few solvent molecules are in contact with surface of solid. However, the rate at which the solvent molecules leave, surface of solid remains unchanged. That is why, temperature is lowered to restore the equalibrium. The freezing depression in a dilute solution is proportional to molality of the solute. The freezing point of benzene solution was 5.4^(@)C . The osmotic pressure of same solution at 10^(@)C is (freezing point of benzene = 5.5^(@)C ). Assume solution to be dilute. [ K_(f) for C_(6)H_(6) is 4.9 K "molality"^(-1) ].

Addition of non-volatile solute to solvent lowers its vapoure pressure. Therefore, the vapour pressure of a solution (i.e, V.P. of solvent in a solution) is lower than that of pure solvent in a solution) is lower than that of pure solvent, at the same temperature. A higher temperature is needed to raise the vapour pressure upto one atmosphere pressure, when boiling point is attined. However, increase in b.pt. is small . for example, 0.1 molal aqueous sucrose solution boils at 10.05^(@)C Sea water, an aqueous solution, which is rich in Na^(+) and Cl^(-) ions, freezes about 1^(@)C lower than frozen water . At the freezing point of a pure-solvent, the reates at which two molecule stick together to form the solid and leave it to return to liquid are equal when solute is present. Few solvent molecules are in contact with surface of solid. However, the rate at which the solvent molecules leave, surface of solid remains unchanged. That is why, temperature is lowered to restore the equalibrium. The freezing depression in a dilute solution is proportional to molality of the solute. When 250 mg of eugonal is added to 100 g of camhor (K_(f) = 29.7 K "molality"^(-1)) . it lowered the freezing point by 0.62^(@)C . The molar mass of eugonal is :

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