Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is

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

When 20 g of naphthoic acid (C₁₁H₈O₂) is dissolved in 50 g of benzene (K_f= 1.72 K kg mol^-1), a freezing point depression of 2 K is observed. The Van’t Hoff factor (i) is

K_f for water is 1.86 K kg mol^-1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol(C₂H₆O₂) must you add to get the freezing point of the solution lowered to –2.8ºC ?

A 5% solution of cane sugar (molar mass 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/mol is :

On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components(heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane= 100 g mol^-1 and of octane = 114 g mol^-1)

Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution,vapour pressure of the solution increases by 10 mm Hg. Vapour pressure ( in mm Hg) of X and Y in their pure states will be, respectively:

The vapour pressure of water at 20°C is 17.5 mm Hg. If 18 g of glucose (C₆H₁₂O₆) is added to 178.2 g of water at 20°C,the vapour pressure of the resulting solution will be

At 80° C, the vapour pressure of pure liquid ‘A’is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80° C and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol^-1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm–3, molar mass of the substance will be