Solution

KNO₃ is a strong electrolyte which dissociates into two ions. Therefore, its van’t Hoff factor is 2. Acetic acid (CH₃COOH) is a weak electrolyte, it does not dissociate completely. So, its van’t Hoff factor is less than that of KNO₃
∴ Osmotic pressure of 0.1 M KNO₃ > Osmotic pressure of 0.1 M CH₃COOH
or P₁ > P₂

Which of the following pairs of solution are isotonic at the same temperature ?

0.1 M Ca(NO₃)₂ and 0.1M Na₂SO₄
0.1 M NaCl and 0.1 M Na₂SO₄
0.1 M urea and 0.1 M MgCl₂
0.2 M urea and 0.1 M NaCl

Solution

274.339 K
– 1.339 K
257.3K
– 1.339°C

Solution

If A contains 2% NaCl and is separated by a semipermeable membrane from B which contains 10% NaCl, then which event will occur ?

NaCl will flow from A to B
NaCl will flow from B to A
Water will flow from A to B
Water will flow from B to A

Solution

Solvent molecules flow from low concentration solution to high concentration solution.

4 gm castic soda is dissolved in 100 cc of solution. The normality of solution is

zero
0.5
1
1.5

Solution

The temperature at which 10% aqueous solution (w/v) of glucose will exhibit the osmotic pressure of 14 bar is –

307.3°C
307.3K
34 K
– 34°C

Solution

The normal boiling point of the solution is the temperature at which the vapour pressure of the solution is –

equal to 1 torr
equal to 76 mm Hg
equal to 2.0 atm
equal to 1 atm

Solution

Boiling temperature is a temperature at which vapour pressure is equal to the atmospheric pressure when external pressure is equal to 1 atm.

The solutions which have same osmotic concentrations are known as

Normal
Isotonic
Hypotonic
Hypertonic

Solution

Isotonic solutions are those solutions which have same osmotic concentrations.The solution with higher osmotic concentration is called hypertonic and lower is called hypotonic solution.

The rise in the boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is 0.1°C. The molal elevation constant of the liquid is

0.01 K/m
0.1 K/m
1 K/m
10 K/m

Solution

K_b = \(\frac{0.1 \times 180 \times 100}{1.8 \times 1000}\) = 1 K/m

A 0.5 molal solution of ethylene glycol in water is used as coolant in a car. If the freezing point constant of water be 1.86°C per mole, the mixture shall freeze at

0.93°C
–0.93°C
1.86°C
–1.86°C

Solution

ΔT_f = K_f × m = 1.86 × 0.5 = 0.93°C ; T_f = -0.93°C

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