The freezing point depression constant format is -1.86°C m^-1. If 5.00g Na₂SO₄ dissolved

in 45.0 g H₂O, the freezing point is changed by -3.82°C. Calculate the van't Hoff factor

for NaSO4

1. 2.63

2. 3.11

3. 0.381

4. 2.05

The van't Hoff factor, i for a compound which undergoes dissociation in one solvent and

association in other solvent is respectively.

1. less than one and less than one

2. greater than one and less than one

3. greater than one and greater than one

4. less than one and greater than one

An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of

the solution to increase ?

1. Addition of NaCl

2. Addition of Na₂SO₄

3. Addition of 1.00 molal KI

4. Addition of water

A solution of sucrose (molar mass = 342 g mol^-1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water.

The freezing point of the solution obtained will be:

 (K_f for water = 1.86 K kg mol^-1)

1. -0.372 $°$C

2. 0.372 $°$C

3. 0.572 $°$C

4. -0.572 $°$C

A 0.0020 m aqueous solution of an ionic compound Co(NH₃)₅(NO₂)Cl freezes at -0.00732°C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (k_f=-1.86^$°$C/ m)

1. 2

2. 3

3. 4

4. 1

0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If K_f for water is 1.86 K kg mol^-1, the lowering in freezing point of the solution is:

(1) -1.12 K

(2) 0.56 K

(3) 1.12 K

(4) - 0.56 K

1.00 g of a non-electrolyte solute (molar mass 250g/mol) was dissolved in 51.2 g of benzene. If the freezing point depression constant, K_f of benzene is 5.12 K kg mol^-1, the freezing point of benzene will be lowered by:-

(1) 0.4 K

(2) 0.3 K

(3) 0.5 K

(4) 0.2 K

A solution of acetone in ethanol:

1. Shows a negative deviation from Raoult's law.

2. Shows a positive deviation from Raoult's law.

3. Behaves like a near ideal solution.

4. Obeys Raoult's law.

Three solutions are prepared by adding 'w' gm of 'A' into 1kg of water, 'w' gm of 'B' into another 1 kg of water and 'w' gm of 'C' in another 1 kg of water (A, B, C are non electrolytic). Dry air is passed from these solutions in sequence (A → B → C). The loss in weight of solution A was found to be 2gm while solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A, B and C is :

(1) MA: M_B : Mc = 4 : 3 : 5

(2) M_A : M_B: M_c = $\frac{1}{4}:\frac{1}{3}:\frac{1}{5}$

(3) Mc > MA > MB

(4) MB > M_A> M_c

How many mili moles of sucrose should be dissolved in 500 gms of water so as to get a solution which has a difference of 103.57°C between boiling point and freezing point.

(K₁ = 1.86 K Kg mol⁻¹, K_b = 0.52 K Kg mo1⁻¹)

(1) 500 mmoles

(2) 900 mmoles

(3) 750 mmoles

(4) 650 mmoles