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 (kf for water = 1.86 K kg mol-1)

(a) -0.372$°$C

(b) 0.372$°$C

(c) 0.572$°$C

(d)  -0.572$°$C

Concept Questions :-

Depression of Freezing Point
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

(a) increase in ionic mobility of ions

(b) 100% ionisation of electrolyte at normal dilution

(c) increase in both, i.e. number of ions and ionic mobility of ions

(d) increase in number of ions

Concept Questions :-

Van’t Hoff Factor
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mho cm2 and at infinite dilution is 400 mho cm2 .The dissociation constant of this acid is.

(a) 1.25x10-5
(b) 1.25x10-6
(c) 6.25x10-4
(d) 1.25x10-4

Concept Questions :-

Van’t Hoff Factor
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

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 (kf=-1.86C/ m)
(a) 2
(b) 3
(c) 4
(d) 1

Concept Questions :-

Depression of Freezing Point
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

Kohlrausch's law states that at

(a) finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

(b) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of th electrolyte

(c) infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte

(d) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

Concept Questions :-

Azeotrope
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Difficulty Level:

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

(a) -1.12 K

(b) 0.56 K

(c) 1.12 K

(d) - 0.56 K

High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

A solution containing 10 g per dm3 of urea (molecular mass = 60 g mol-1) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is:

(a) 250 g mol-1                (b) 300 g mol-1

(c) 350 g mol-1                (d) 200 g mol-1

Concept Questions :-

Concentration Term and Numericals
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

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, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by:-

(a) 0.4 K
(b) 0.3 K
(c) 0.5 K
(d) 0.2 K

Concept Questions :-

Depression of Freezing Point
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

A solution of acetone in ethanol:

(a) shows a negative deviation from Raoult's law

(b) shows a positive deviation from Raoult's law

(c) behaves like a near ideal solution

(d) obeys Raoult's law

Concept Questions :-

Introduction/ Colligative properties
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

During osmosis, flow of water through a semi-permeable membrane is :

(a) from solution having higher concentration only

(b) from both sides of semi-permeable membrane with equal flow rates

(c) from both sides of semi-permeable membrane with unequal flow rates

(d) from solution having lower concentration only

Concept Questions :-

Osmosis and Osmotic Pressure