Chapter 2 Solutions Solutions
Question - 21 : - A sample of drinking water wasfound to be severely contaminated with chloroform (CHCl3) supposedto be a carcinogen. The level of contamination was 15 ppm (by mass):
Answer - 21 : -
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.
Answer
Question - 22 : - What role does the molecularinteraction play in a solution of alcohol and water?
Answer - 22 : -
In pure alcohol and water, themolecules are held tightly by a strong hydrogen bonding. The interactionbetween the molecules of alcohol and water is weaker than alcohol−alcohol andwater−water interactions. As a result, when alcohol and water are mixed, theintermolecular interactions become weaker and the molecules can easily escape.This increases the vapour pressure of the solution, which in turn lowers theboiling point of the resulting solution.
Question - 23 : - Why do gases always tend to beless soluble in liquids as the temperature is raised?
Answer - 23 : -
Solubility of gases in liquidsdecreases with an increase in temperature. This is because dissolution of gasesin liquids is an exothermic process.
Therefore,when the temperature is increased, heat is supplied and the equilibrium shiftsbackwards, thereby decreasing the solubility of gases.
Question - 24 : - State Henry’s law and mentionsome important applications?
Answer - 24 : -
Henry’s law states thatpartial pressure of a gas in the vapour phase is proportional to the molefraction of the gas in the solution. If p isthe partial pressure of the gas in the vapour phase and x isthe mole fraction of the gas, then Henry’s law can be expressed as:
p = KH x
Where,
KH isHenry’s law constant
Some important applications ofHenry’s law are mentioned below.
(i) Bottlesare sealed under high pressure to increase the solubility of CO2 insoft drinks and soda water.
(ii) Henry’slaw states that the solubility of gases increases with an increase in pressure.Therefore, when a scuba diver dives deep into the sea, the increased seapressure causes the nitrogen present in air to dissolve in his blood in greatamounts. As a result, when he comes back to the surface, the solubility ofnitrogen again decreases and the dissolved gas is released, leading to theformation of nitrogen bubbles in the blood. This results in the blockage ofcapillaries and leads to a medical condition known as ‘bends’ or ‘decompressionsickness’.
Hence, the oxygen tanks used byscuba divers are filled with air and diluted with helium to avoid bends.
(iii) Theconcentration of oxygen is low in the blood and tissues of people living athigh altitudes such as climbers. This is because at high altitudes, partialpressure of oxygen is less than that at ground level. Low-blood oxygen causesclimbers to become weak and disables them from thinking clearly. These aresymptoms of anoxia.
Question - 25 : - The partial pressure of ethaneover a solution containing 6.56 × 10−3 g of ethane is 1 bar. Ifthe solution contains 5.00 × 10−2 g of ethane, then what shallbe the partial pressure of the gas?
Answer - 25 : - 
Number of moles present in 5.00 × 10−2 gof ethane 
= 1.67 × 10−3 mol
According to Henry’s law,
p = KHx
= 7.636 bar
Hence, partial pressure of thegas shall be 7.636 bar.
What is meant by positive andnegative deviations from Raoult's law and how is the sign of ΔsolH relatedto positive and negative deviations from Raoult's law?
Answer - 26 : -
According to Raoult’s law, thepartial vapour pressure of each volatile component in any solution is directlyproportional to its mole fraction. The solutions which obey Raoult’s law overthe entire range of concentration are known as ideal solutions. Thesolutions that do not obey Raoult’s law (non-ideal solutions) have vapourpressures either higher or lower than that predicted by Raoult’s law. If thevapour pressure is higher, then the solution is said to exhibit positivedeviation, and if it is lower, then the solution is said to exhibit negative deviationfrom Raoult’s law.
Vapour pressure of a two-component solution showingpositive deviation from Raoult’s law
Vapour pressure of a two-component solution showingnegative deviation from Raoult’s law
In the case of an ideal solution,the enthalpy of the mixing of the pure components for forming the solution iszero.
ΔsolH = 0
In the case of solutions showingpositive deviations, absorption of heat takes place.
∴ΔsolH = Positive
In the case of solutions showingnegative deviations, evolution of heat takes place.
∴ΔsolH = Negative
Question - 27 : - An aqueous solution of 2%non-volatile solute exerts a pressure of 1.004 bar at the normal boiling pointof the solvent. What is the molar mass of the solute?
Answer - 27 : -
Here,
Vapour pressure of the solutionat normal boiling point (p1) = 1.004 bar
Vapourpressure of pure water at normal boiling point 
Mass of solute, (w2)= 2 g
Mass of solvent (water), (w1)= 98 g
Molar mass of solvent (water), (M1)= 18 g mol−1
According to Raoult’s law,
= 41.35 g mol−1
Hence, the molar mass of thesolute is 41.35 g mol−1.
Heptane and octane form an idealsolution. At 373 K, the vapour pressures of the two liquid components are 105.2kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of26.0 g of heptane and 35 g of octane?
Answer - 28 : - Vapour pressure of heptane
Vapour pressure of octane
= 46.8 kPa We know that,
Molar mass of heptane (C7H16)= 7 × 12 + 16 × 1
= 100 g mol−1
Number of moles ofheptane 
= 0.26 molMolar mass of octane (C8H18)= 8 × 12 + 18 × 1
= 114 g mol−1
Number of moles of octane
= 0.31 mol
Molefraction of heptane,
= 0.456
And, mole fraction ofoctane, x2 = 1 − 0.456
= 0.544
Now,partial pressure of heptane, 
= 0.456 × 105.2
= 47.97 kPa
Partialpressure of octane, 
= 0.544 × 46.8
= 25.46 kPa
Hence, vapour pressure ofsolution, ptotal = p1 + p2
= 47.97 + 25.46
= 73.43 kPa
The vapour pressure of water is12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of anon-volatile solute in it.
Answer - 29 : -
1 molal solution means 1 mol ofthe solute is present in 1000 g of the solvent (water).
Molar mass of water = 18 g mol−1
Number of molespresent in 1000 g of water
= 55.56 mol
Therefore, mole fraction of thesolute in the solution is
It is given that,
Vapour pressure of water, 
= 12.3 kPa
Applyingthe relation, 
⇒ 12.3 − p1 = 0.2177
⇒ p1 = 12.0823
= 12.08 kPa (approximately)
Hence, the vapour pressure of thesolution is 12.08 kPa.
Calculate the mass of anon-volatile solute (molar mass 40 g mol−1) which should bedissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer - 30 : - Let the vapour pressure of pure octane be
Then, the vapour pressure of the octane afterdissolving the non-volatile solute is
Molar mass of solute, M2 =40 g mol−1
Mass of octane, w1 =114 g
Molar mass of octane, (C8H18), M1 =8 × 12 + 18 × 1
= 114 g mol−1
Applyingthe relation,Hence, the required mass of thesolute is 8 g.