Which of the following solutions will have a lower vapour pressure and why? (i) A 5 % solution of cane sugar ( C 12 H 22 O 11 ) (ii) A 5 % solution of urea ( NH 2 CONH 2 ) (Relative atomic masses of H = 1 , C = 12 , O = 16 , N = 14 )

Lowering of vapour pressure of a solution is directly proportional to the number of moles of solute present in the solution. Greater the number of moles of solute present, lower will be the vapour pressure or greater will be the extent of lowering of vapour pressure of that solution. Number of moles of solute present in the solution = mass of solute in grams molecular mass of solute in g · mol - 1 (i) A 5 % solution of cane sugar ( C 12 H 22 O 11 ) will have: Molecular mass = ( 12 × 12 ) + 22 + ( 16 × 11 ) = 342 g / mol Mass of solute, cane sugar = 5 g Number of moles of cane sugar = 5 342 = 0 . 014 (ii) A 5 % solution of urea ( NH 2 CONH 2 ) will have: Molecular mass = ( 2 × 14 ) + 4 + 12 + 16 = 60 g / mol Mass of solute, urea = 5 g Number of moles of urea = 5 60 = 0 . 083 As the numerical value of number of moles of urea is greater than that of cane sugar, hence, the 5 % solution of urea ( NH 2 CONH 2 ) will have comparatively lower vapour pressure.

E M B I B E

ICSE Chemistry>Chapter 2 - Solutions>QUESTIONS FROM ISC EXAMINATION PAPERS>Q 7

HARD

12th ICSE

IMPORTANT

Earn 100

The boiling point of pure water is $373 K$ . Calculate the boiling point of an aqueous solution containing $18 gms$ of glucose (MW $= 180$ ) in $100 gms$ of water. Molal elevation constant of water is $0.52 K \cdot Kg \cdot {mol}^{- 1}$ .

Important Questions on Solutions

HARD

12th ICSE

IMPORTANT

ICSE Chemistry>Chapter 2 - Solutions>QUESTIONS FROM ISC EXAMINATION PAPERS>Q 7

Equal weights of two substances X and Y are dissolved in equal volumes of water. The osmotic pressure of the solution containing Y is five times the osmotic pressure of the solution containing X. What is the molecular weight of X if that of Y is