Chapter 2 Solutions Solutions
Question - 31 : - A solution containing 30 g ofnon-volatile solute exactly in 90 g of water has a
vapour pressure of 2.8 kPa at298 K.
Answer - 31 : -
Further, 18 g of water is then added to
the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
i. molar mass of the solute
ii. vapour pressure of water at 298 K.
Answer
(i) Let,the molar mass of the solute be M g mol−1
Now,the no. of moles of solvent (water), 
And, the no. of moles of solute,
Applying the relation:
After the addition of 18 g ofwater:
Again, applying the relation:
Dividing equation (i) by (ii),we have:
Therefore, the molar mass of thesolute is 23 g mol−1.
(ii) Putting the value of ‘M’ in equation (i),we have:
Hence, the vapour pressure ofwater at 298 K is 3.53 kPa.
Question - 32 : - Two elements A and B formcompounds having formula AB2 and AB4. When dissolvedin 20 g of benzene (C6H6),
Answer - 32 : - 1 g of AB_2 lowers the freezing point by 2.3 Kwhereas 1.0 g of AB_4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol_−1. Calculate atomic masses of A and B.
Answer
Now, we can write:
Subtracting equation (i) from(ii), we have
2y = 85.28
⇒ y = 42.64
Putting the value of ‘y’in equation (1), we have
x + 2× 42.64 = 110.87
⇒ x = 25.59
Hence, the atomic masses of A andB are 25.59 u and 42.64 u respectively.
At 300 K, 36 g of glucose presentin a litre of its solution has an osmotic pressure of 4.98 bar. If the osmoticpressure of the solution is 1.52 bars at the same temperature, what would beits concentration?
Answer - 33 : -
T =300 K
π = 1.52 bar
R = 0.083 bar L K−1 mol−1
Applying the relation,
π = CRT
= 0.061 mol
Since the volume of the solutionis 1 L, the concentration of the solution would be 0.061 M.
Question - 34 : - Suggest the most important type of intermolecular attractive interaction in the following pairs.
Answer - 34 : -
(i) n-hexaneand n-octane
(ii) I2 andCCl4
(iii) NaClO4 andwater
(iv) methanoland acetone
(v) acetonitrile(CH3CN) and acetone (C3H6O).
Answer
(i) Van derWall’s forces of attraction.
(ii) Van derWall’s forces of attraction.
(iii) Ion-diopleinteraction.
(iv) Dipole-dipoleinteraction.
(v) Dipole-dipoleinteraction.
Based on solute-solventinteractions, arrange the following in order of increasing solubility inn-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Answer - 35 : -
n-octane is a non-polar solvent.Therefore, the solubility of a non-polar solute is more than that of a polarsolute in the n-octane.
The order of increasing polarityis:
Cyclohexane < CH3CN< CH3OH < KCl
Therefore, the order ofincreasing solubility is:
KCl < CH3OH < CH3CN< Cyclohexane
Question - 36 : - Amongst the following compounds,identify which are insoluble, partially soluble and highly soluble in water?
Answer - 36 : -
(i) phenol (ii) toluene (iii) formic acid
(iv) ethylene glycol (v) chloroform (vi) pentanol.
Answer
(i) Phenol(C6H5OH) has the polar group −OH and non-polar group −C6H5.Thus, phenol is partially soluble in water.
(ii) Toluene(C6H5−CH3) has no polar groups. Thus, tolueneis insoluble in water.
(iii) Formicacid (HCOOH) has the polar group −OH and can form H-bond with water. Thus,formic acid is highly soluble in water.
(iv) Ethyleneglycol 
has polar −OH group and canform H−bond. Thus, it is highly soluble in water.
(v) Chloroformis insoluble in water.
(vi) Pentanol(C5H11OH) has polar −OH group, but it also contains avery bulky non-polar −C5H11 group. Thus, pentanolis partially soluble in water.
Question - 37 : - If the density of some lake wateris 1.25 g mL−1 and contains 92 g of Na+ ionsper kg of water, calculate the molality of Na+ ions in thelake.
Answer - 37 : - Number of moles present in 92 g of Na+ ions = 
= 4 mol
Therefore,molality of Na+ ions in the lake 
= 4 m
Question - 38 : - If the solubility product of CuSis 6 × 10−16, calculate the maximum molarity of CuS in aqueoussolution.
Answer - 38 : -
Solubility product of CuS, Ksp =6 × 10−16
Let s be thesolubility of CuS in mol L−1.
Now,
= s × s
= s2
Then,we have, Ksp = 
= 2.45 × 10−8 molL−1
Hence, the maximum molarity ofCuS in an aqueous solution is 2.45 × 10−8 mol L−1.
Question - 39 : - Calculate the mass percentage ofaspirin (C9H8O4) in acetonitrile (CH3CN)when 6.5 g of C9H8O4 is dissolved in 450g of CH3CN.
Answer - 39 : -
.5 g of C9H8O4 isdissolved in 450 g of CH3CN.
Then, total mass of the solution= (6.5 + 450) g
= 456.5 g
Therefore,mass percentage ofC9H8O4 
Question - 40 : - Nalorphene (C19H21NO3),similar to morphine, is used to combat withdrawal
Answer - 40 : -
symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg.
Calculate the mass of 1.5 × 10_−3m aqueous solution required for the above dose.
Answer
The molar mass of nalorphene 
is given as:
In 1.5 × 10−3m aqueoussolution of nalorphene,
1kg (1000 g) of water contains 1.5 × 10−3 mol
Therefore, total mass of the solution 
This implies that the massof the solution containing 0.4665 g of nalorphene is 1000.4665 g.
Therefore, mass of the solutioncontaining 1.5 mg of nalorphene is:
Hence, the mass of aqueoussolution required is 3.22 g.