1.00 g of non–electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by
|
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase? |
Answer |
|
The molarity of a NaOH solution by dissolving 4 g of it in 250 ml water is |
Answer |
|
What is the osmotic pressure of a 0.0020 mol dm–3 sucrose (C12H22O11) solution at 20°C? (Molar gas constant, R = 8.314 JK–1 mol–1, 1 dm3 = 0.001 m3? |
Answer |
|
The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is |
Answer |
|
0.5 molal aqueous solution of a weak acid (HX) is 20% ionized. If Kf for water is 1.86 K kg mol–1, the lowering in freezing point of the solution is |
Answer |
|
pA and pB are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be |
Answer |
|
Benzene and toluene form nearly ideal solutions. At 20°C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20°C for a solution containing 78 g of benzene and 46 g of toluene in torr is |
Answer |
|
6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration ofsolution is |
Answer |
|
A solution of sucrose (molar mass = 342 g mol–1) has been prepared by dissolving 68.5 g of sucrose in 100 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol–1) |
Answer |
|
A 5% solution of cane sugar (molar mass 342 g mol–1) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/mol is |
Answer |