Chapter 2 Solutions Solutions
Question - 11 : - Calculate the mass of ascorbicacid (Vitamin C, C6H8O6) to be dissolved in
75 g of acetic acid to lower its melting point by 1.5°C. Kf =3.9 K kg mol−1.
Answer - 11 : -
of acetic acid, w1 =75 g
Molar mass of ascorbic acid (C6H8O6), M2 =6 × 12 + 8 × 1 + 6 × 16
= 176 g mol−1
Lowering of melting point, ΔTf =1.5 K
We know that:
= 5.08 g (approx)
Hence, 5.08 g of ascorbic acid isneeded to be dissolved.
Question - 12 : - Calculate the osmotic pressure inpascals exerted by a solution prepared by dissolving 1.0 g of polymer of molarmass 185,000 in 450 mL of water at 37°C.
Answer - 12 : -
It is given that:
Volume of water, V =450 mL = 0.45 L
Temperature, T =(37 + 273)K = 310 K
Numberof moles of the polymer,
We know that:
Osmoticpressure,
= 30.98 Pa
= 31 Pa (approximately)
Define the term solution. Howmany types of solutions are formed? Write briefly about each type with anexample.
Answer - 13 : -
Homogeneous mixtures of two ormore than two components are known as solutions.
There are three types ofsolutions.
(i) Gaseous solution:
The solution in which the solventis a gas is called a gaseous solution. In these solutions, the solute may beliquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is agaseous solution.
(ii) Liquid solution:
The solution in which the solventis a liquid is known as a liquid solution. The solute in these solutions may begas, liquid, or solid.
For example, a solution ofethanol in water is a liquid solution.
(iii) Solid solution:
The solution in which the solventis a solid is known as a solid solution. The solute may be gas, liquid orsolid. For example, a solution of copper in gold is a solid solution.
Question - 14 : - Give an example of solid solutionin which the solute is a gas.
Answer - 14 : -
In case a solid solution isformed between two substances (one having very large particles and the otherhaving very small particles), an interstitial solid solution will be formed.For example, a solution of hydrogen in palladium is a solid solution in whichthe solute is a gas.
Question - 15 : - Define the following terms:
Answer - 15 : -
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Answer
Question - 16 : - Concentrated nitric acid used inlaboratory work is 68% nitric acid by mass in aqueous solution. What should bethe molarity of such a sample of the acid if the density of the solution is1.504 g mL−1?
Answer - 16 : - 
Molarity of solution 
Question - 17 : - A solution of glucose in water islabelled as 10% w/w, what would be the molality and mole fraction of eachcomponent in the solution? If the density of solution is 1.2 g mL−1,then what shall be the molarity of the solution?
Answer - 17 : -
10% w/w solution of glucose inwater means that 10 g of glucose in present in 100 g of the solution i.e., 10 gof glucose is present in (100 − 10) g = 90 g of water.
Molar mass of glucose (C6H12O6)= 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol−1
Then,number of moles of glucose
Question - 18 : - How many mL of 0.1 M HCl arerequired to react completely with 1 g mixture of Na2CO3 andNaHCO3 containing equimolar amounts of both?
Answer - 18 : - 
Hence, 159 mL of 0.1 M of HCl isrequired to react completely with 1 g mixture of Na2CO3 andNaHCO3, containing equimolar amounts of both.
A solution is obtained by mixing300 g of 25% solution and 400 g of 40% solution by mass. Calculate the masspercentage of the resulting solution.
Answer - 19 : - 
Question - 20 : - An antifreeze solution isprepared from 222.6 g of ethylene glycol (C2H6O2)and 200 g of water. Calculate the molality of the solution. If the density ofthe solution is 1.072 g mL−1, then what shall be the molarity of thesolution?
Answer - 20 : - Molar mass of ethylene glycol
= 2 × 12 + 6 × 1 + 2 ×16
= 62 gmol−1
Numberof moles of ethylene glycol
= 3.59 mol
Therefore,molality of the solution = 17.95 m
Total mass of the solution =(222.6 + 200) g
= 422.6 g
Given,
Density of the solution = 1.072 gmL−1
Volume of the solution
= 394.22 mL
= 0.3942 × 10−3 L
Molarity of thesolution 