Solutions Exercise Question & Answers(Q2.22-Q2.41)
- By srinivaseducation
- On 03/02/2013
SOLUTIONS QUESTION AND ANSWERS (Q2.23-Q2.41)
Q2.23:- Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O).
Ans;- (i) Van der Wall’s forces of attraction.
(ii) Van der Wall’s forces of attraction.
(iii) Ion-diople interaction.
(iv) Dipole-dipole interaction.
(v) Dipole-dipole interaction.
Q2.24:- Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Ans:- n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane.
The order of increasing polarity is:
Cyclohexane < CH3CN < CH3OH < KCl
Therefore, the order of increasing solubility is:
KCl < CH3OH < CH3CN < Cyclohexane
Q2.25:- Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid
(iv) ethylene glycol (v) chloroform (vi) pentanol.
Ans:- (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, toluene is insoluble in water.
(iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water. Thus, formic acid is highly soluble in water.
(iv) Ethylene glycol has polar −OH group and can form H−bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water.
(vi) Pentanol (C5H11OH) has polar −OH group, but it also contains a very bulky non-polar −C5H11 group. Thus, pentanol is partially soluble in water.
Q2.26:- If the density of some lake water is 1.25 g mL−1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.
Ans:- Number of moles present in 92 g of Na+ ions =
= 4 mol
Therefore, molality of Na+ ions in the lake
= 4 m
Q2.27:- If the solubility product of CuS is 6 × 10−16, calculate the maximum molarity of CuS in aqueous solution.
Ans:- Solubility product of CuS, Ksp = 6 × 10−16
Let s be the solubility of CuS in mol L−1.
= s × s
Then, we have, Ksp =
= 2.45 × 10−8 mol L−1
Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10−8 mol L−1.
Q2.28:- Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
Ans:- 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
Then, total mass of the solution = (6.5 + 450) g
= 456.5 g
Therefore, mass percentage ofC9H8O4
Q2.29:- Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal
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.
Ans:- The molar mass of nalorphene is given as:
In 1.5 × 10−3m aqueous solution of nalorphene,
1 kg (1000 g) of water contains 1.5 × 10−3 mol
Therefore, total mass of the solution
This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.
Therefore, mass of the solution containing 1.5 mg of nalorphene is:
Hence, the mass of aqueous solution required is 3.22 g.
Q2.30:- Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Ans:- 0.15 M solution of benzoic acid in methanol means,
1000 mL of solution contains 0.15 mol of benzoic acid
Therefore, 250 mL of solution contains = mol of benzoic acid
= 0.0375 mol of benzoic acid
Molar mass of benzoic acid (C6H5COOH) = 7 × 12 + 6 × 1 + 2 × 16
= 122 g mol−1
Hence, required benzoic acid = 0.0375 mol × 122 g mol−1
= 4.575 g