Competition Level Test: Solutions | Physical Chemistry for NEET PDF Download

Larger the number of species in solution, larger the boiling point. 0.01 M Na₂SO₄ solution has 0.03 M of species

Concentration of urea solution,

The molecular interactions between benzene and methanol are weaker than those existing between benzene and benzene, and methanol and methanol. Consequently, the solution will exhibit positive deviations from Raoult’s law.

The freezing point depression is given by –ΔT_f = K_f m. Different solvents have different values of K_f and hence show different freezing point depression.

Na₂SO₄ ionizes as

Total amount of species in solution will be n' = n(1– α) + 2nα + nα = n(1+ 2α)
The van’t Hoff factor is

Amount of benzene,

Amount of toluene,

Amount fraction of benzene,

Partial pressure of benzene,

The solutions will have the same boiling point and freezing point provided the solutes dissolved are nonelectrolytes.

Both acetonitrile and acetone are polar and have permanent dipole moment.

Vapour pressure of water in solution at 100°C is p = xp* = (0.99) (760 Torr) = 752.40 Torr

Isotonic solutions have the same osmotic pressure. Hence, from

c₁ = c₂ ⇒ n₁ = n₂

i.e. M = 210 g mol^–1

For

we have


Mass of AgIO₃ in 100 mL solution is
m = (1.0 x 10^–4 mol L^–1) (283 g mol^–1) (0.1 L) = 2.83 x 10³ g

Using the expression p = x_A p_A + x_B p_B; (A is ethyl alcohol and B is propyl alcohol)

we get  290 mmHg = 0.6 p_A + 0.4 x 200 mmHg

This give  

Let x_A be the amount fraction of A in the liquid solution. We will have

This gives 

Hence, the amount per cent of A is

Amount of glucose,

Amount of water,

Mole fraction of water,

Vapour pressure of solution,

 For an ideal solution

The interaction between unlike molecules is weaker than those involved between like molecules. This results into positive deviations from Raoult’s law

van't Hoff factor = 3
Hence, 
– ΔT_f  = i K_f m = (3) (1.86 K kg mol^–1) (0.01 mol kg^–1) = 0.0558 K

Amount of heptane, 

Amount of octane,

Mole fraction of heptane,

Mole fraction of octane, x₂ = 1 – x_x = 0.55
According to Raoult’s law

= (0.45) (105 kPa) + (0.55) (45 kPa) = 72.0 kPa

Since

we have

We have

Total molality of the solution = m(1 – α) + m(xα) + m(yα) = m{(x + y – l) α + 1}
The van’t Hoff factor is 

or i = (x + y – 1)α + 1 i.e.

In 5.2 molal aqueous solution, we have
n₂ = 5.2 mol and n₁ = (1000 g/18 g mol^–1) = 55.56 mol
The mole fraction of methyl alcohol in the solution is

Isotonic solutions have identical osmotic pressure and hence identical concentrations, i.e. the same amounts in the fixed volume of solutions. Thus, 

This gives

Assuming density of water equal to 1 kg dm^–3, we have

Since – ΔT_f = K_f m, we get.

Since the molality, m = n/m₁ , we get
n = mm₁ = (1.505 mol kg^–1) (1 kg) = 1.505 mol  
Finally, the mass of ethylene glycol, C₂H₆O₂ (the molar mass = 62 g mol^–1) required will be m = n
M_m = (1.505 mol)(62 g mol^–1) = 93.3 g

All the four solutions have the same amount of the species present in the solution.

Since all of them have the concentration of total species, their osmotic pressure will have the same value.