Important questions based on NCERT syllabus for Chapter 2 - Solutions:
Question-1: Calculate the molarity of each of the following solutions: (a) 30 g of CO(NO3)2.6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
Solution: (a) Molar mass of CO(NO3)2.6H2O=310.7 g mol-1
no. of moles = 30/310.7 = 0.0966
Vol. of solution = 4.3 L
Molarity =0.0966/4.3 = 0.022M
(b) 1000 mL of 0.5M H2SO4 contain H2SO4 = 0.5 mole
30 mL of 0.5 M H2SO4 contain H2SO4
=0.5/1000 x 30 = 0.015 mole
Volume of solution = 500mL=0.5 L
Molarity = 0.015/0.5 = 0.03M
Question-2: Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
Solution: 0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH2CONH2) = 60 g mol-1
.’. 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
.’. 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g
Question-3: What role does the molecular interaction play in a solution of alcohol and water?
Solution: Alcohol and water both have strong tendency to form intermolecular hydrogen bonding. On mixing the two, a solution is formed as a result of formation of H-bonds between alcohol and H2O molecules but these interactions are weaker and less extensive than those in pure H2O. Thus they show a positive deviation from ideal behaviour. As a result of this, the solution of alcohol and water will have higher vapour pressure and lower boiling point than that of water and alcohol.