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Addition of non-volatile solute to solve...

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 amount of ice seperated out on cooling a solution containing `50g` ethylene glycol in `200g` water to `-9.3^(@)C` is : [`K_(f)` for `H_(2)O = 1.86 K "molality"^(-1)`]

A

`38.71 g`

B

`61.29 g`

C

`138.71 g`

D

`161.29 g`

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The correct Answer is:
A
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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 a solution containing 50 cm^(3) of ethylene glycol in 50g water is found to be -34^(@)C . Density of ethylene glycol is : Assuming solution is dilute [ K_(f) for H_(2)O = 1.86 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. The freezing point of a 5g CH_(3)COOH(aq) per 100 g water is -1.576^(@)C . Then van't Hoff factor ( K_(f) of water = 1.86 K "mol"^(-1) kg ) :

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 :

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. 2g of benzoic acid dissolved in 25g of C_(6)H_(6) shows a depression in f.pt, equal to 1.62 K. K_(f) for C_(6)H_(6) in 4.9 K "molality"^(-1) . The percentage of dimerisation is :