The vapour pressure of 1 molal solution of a non-volatile solute in water at 300 K would be -

(The vapour pressure of water at 300 K = 12.3 kPa .)

A solution containing 30 g of non-volatile solute in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. The molar mass of the solute will be -

Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by  2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The atomic masses of A and B are respectively -

(K_f for benzene is 5.1 K kg mol^-1)

The type of inter molecular interactions present in

(i) n-hexane and n-octane

(ii) NaClO₄ and water

are respectively -

The insoluble compound in water is /are

If the solubility product of CuS is 6 × 10^–16, the maximum molarity of CuS in aqueous solution will be -

The mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN will be -

The dose of nalorphene (C₁₉H₂₁NO₃) is 1.5 mg. The mass of 1.5$\times$10^–3 m aqueous solution required for the above dose is

The amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol is -

The depression in the freezing point of water was observed for the same amount of acetic acid, trichloroacetic acid, and trifluoroacetic in the order of-

The depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water will be -

3. 0.65 K

19.5 g of CH₂FCOOH is dissolved in 500 g of water. The van’t Hoff factor and dissociation constant of fluoroacetic acid will be respectively -

($∆{\mathrm{T}}_{\mathrm{f}}$ = 1.00 K )

100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B

(molar mass 180 g mol^–1). The vapour pressure of pure liquid B was found to be 500 torr. If the total vapour pressure of the solution is 475 torr, the vapour pressure of pure liquid A will be -

The major component of air is nitrogen with approximately 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K. if the Henry’s law constant for nitrogen at 298 K is 6.51 × 10⁷ mm respectively, then the composition of nitrogen in air will be

The amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C will be -

If 22g of benzene is dissolved in 122g of CCl₄ , the mass percentage of CCl₄ and benzene respectively are-

The type of intermolecular interaction present in 

(i) Methanol and acetone

(ii) Acetonitrile and acetone-

The mole fraction of benzene in solution containing 30% by mass in CCl₄ is-

To make 2.5 kg of 0.25 m aqueous solution, the mass of urea required is-

If 1.202 g mL^-1 is the density of 20% aqueous Kl , the molarity of the solution is-

The solubility of H₂S in water at STP is 0.195 m, the value of Henry's constant is-

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. The mass percentage of the solvent in resulting solution is:

The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. The composition of the liquid mixture if total vapour pressure is 600 mm Hg is:

$$\begin{gathered} 1. {\mathrm{x}}_{\mathrm{A}} =0.2 ; {\mathrm{x}}_{\mathrm{B} }= 0.8 \\ 2. {\mathrm{x}}_{\mathrm{A}} = 0.8 ; {\mathrm{x}}_{\mathrm{B}} = 0.2 \\ 3. {\mathrm{x}}_{\mathrm{A} }= 0.4 ; {\mathrm{x}}_{\mathrm{B}} = 0.6 \\ 4. {\mathrm{x}}_{\mathrm{A} }= 0.6 ; {\mathrm{x}}_{\mathrm{B} }= 0.4 \end{gathered}$$

The vapour pressure of water in the solution having 50 g of urea dissolved in 850 g of water is-

(Vapor pressure of pure water at 298 K is 23.8 mm Hg) 

The amount of sucrose is to be added to 500 g of water such that it boils at 100°C is:

( Given: The boiling point of water at 750 mm Hg is 99.63°C ; K_b = 0.52 K kg mol^-1)

The solubility of a gas in a liquid decreases with an increase in temperature because-

To minimise the painful effects accompanying deep sea diving, oxygen diluted with less soluble helium gas is used as breathing gas by the divers. This is an example of the application of- 

The correct match for the above table is -

The example of gas in solid type solution is-

The incorrect formula among the following is -

$$\begin{gathered} 1. {\mathrm{X}}_{\mathrm{mole} \mathrm{fraction}}=\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{n}}_{\mathrm{solution}}} \\ 2. \mathrm{Molarity} = \frac{\mathrm{amount} \mathrm{of} \mathrm{solute} \left(\mathrm{g}\right)}{\mathrm{volume }\mathrm{of} \mathrm{solution} \left(\mathrm{mL}\right)} \\ 3. \mathrm{Molality} = \frac{\mathrm{Number} \mathrm{of} \mathrm{mole} \mathrm{of} \mathrm{solute}}{\mathrm{amount} \mathrm{of} \mathrm{solvent} \left(\mathrm{Kg}\right)} \\ 4. \mathrm{Mass} \mathrm{percentage} = \\ \frac{\mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{component} \mathrm{in} \mathrm{the} \mathrm{solution}}{\mathrm{Total} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{solution}}\times 100 \end{gathered}$$