Solutions

A solution is prepared by adding 1 mole ethyl alcohol in 9 mole water. The mass percent of solute in the solution is (Integer answer) (Given : Molar mass in g mol-1 Ethyl alcohol : 46 water : 18)

Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is x 10-3, Value of x is . (Integer answer)

Consider the dissociation of the weak acid HX as given below HX(aq) H+(aq) + X-(aq), Ka = 1.2 10-5 [Ka : dissociation constant] The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is 10-2 bar (nearest integer). [Given : R = 0.083 L bar mol-1 K-1]

When 'x' 10-2 mL methanol (molar mass = 32 g, density = 0.792 g/cm3) is added to 100 mL water (density = 1 g/cm3), the following diagram is obtained. [figure] x = (nearest integer). [Given : Molal freezing point depression constant of water at 273.15 K is 1.86 K kg mol-1]

An artificial cell is made by encapsulating 0.2 M glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a 0.05 M solution of NaCl at 300 K is 10-1 bar. (nearest integer). [Given : R = 0.083 L bar mol-1 K-1] Assume complete dissociation of NaCl

Considering acetic acid dissociates in water, its dissociation constant is 6.25 10-5. If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at -x 10-2 C, provided pure water freezes at 0 C. x = . (Nearest integer) Given : (Kf)water = 1.86 K kg mol-1 density of acetic acid is 1.2 g mL-1. molar mass of water = 18 g mol-1. molar mass of acetic acid = 60 g mol-1. density of water = 1 g cm-3 Acetic acid dissociates as CH3COOH CH3COO + H

2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25 . The solution showed a boiling point elevation by 2 . Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is mm of Hg (nearest integer) [Given : Molal boiling point elevation constant of water (Kb) = 0.52 K kg mol-1, 1 atm pressure = 760 mm of Hg, molar mass of water = 18 g mol-1]

2.7 kg of each of water and acetic acid are mixed. The freezing point of the solution will be -x C. Consider the acetic acid does not dimerise in water, nor dissociates in water. x = (nearest integer) [Given: Molar mass of water = 18 g mol-1, acetic acid = 60 g mol-1 Kf H2O : 1.86 K kg mol-1 Kf acetic acid: 3.90 K kg mol-1 freezing point: H2O = 273 K, acetic acid = 290 K]

We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentration 0.1 M, 0.01 M and 0.001 M, respectively. The value of van 't Hoff factor (i) for these solutions will be in the order :

1. iA < iC < iB
2. iA < iB < iC
3. iA > iB > iC
4. iA = iB = iC

Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24 is kg (Molar mass in g mol-1 for ethylene glycol 62, Kf of water = 1.86 K kg mol-1)