Samacheer Class 11 Chemistry Solutions Book Back Answers & Solutions

Your Progress - Chapter 90% complete

1I. Choose the best answer:30 questions

Q.1The molality of a solution containing 1.8 g of glucose dissolved in 250 g of water is a) 0.2 M b) 0.01 M c) 0.02 M d) 0.04 Mv

Solution

d) 0.04 M

Answer:

d) 0.04 M

Q.2Which of the following concentration terms is / are independent of temperature a) molality b) molarity c) mole fraction d) a and bv

Solution

d) a and b

Answer:

d) a and b

Q.3Stomach acid, a dilute solution of HCl can be neutralized by reaction with aluminium hydroxide Al(OH) 3 + 3HCl (aq) -> AlCl 3 + 3H 2 O. How many milliliters of 0.1 M Al(OH) 3 solution is needed to neutralize 21 ml of 0.1 M HCl? a) 14 mL b) 7 mL c) 21 mL d) none of thesev

Solution

b) 7 mL

Answer:

b) 7 mL

Q.4The partial pressure of nitrogen in air is 0.76 atm and its Henry’s law constant is 7.6 × 10 4 atm at 300 K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300 K ? a) 1 × 10 -4 b) 1 × 10 4 c) 2 × 10 -5 d) 1 × 10 -5v

Solution

d) 1 × 10 -5

Answer:

d) 1 × 10 -5

Q.5Henry’s law constant for the solubility of Nitrogen gas in water at 350 K is 8 × 10 4 atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressure is a) 4 × 10 -4 b) 4 × 10 4 c) 2 × 10 -2 d) 2.5 × 10 -4v

Solution

d) 2.5 × 10 -4

Answer:

d) 2.5 × 10 -4

Q.6Which one of the following is incorrect for an ideal solution? a) ∆H mix = 0 b) ∆U mix =0 c) ∆P = P observed – P calculated by Raoults law = 0 d) ∆G mix = 0v

Solution

d) ∆G mix = 0

Answer:

d) ∆G mix = 0

Q.7Which one of the following gases has the lowest value of Henry’s law constant? a) N 2 b) He c) CO 2 d) H 2v

Solution

c) CO 2

Answer:

c) CO 2

Q.8P 1 and P 2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution If x 1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be a) P 1 + x 1 (P 2 – P 1 ) b) P 2 – x 1 (P 2 + P 1 ) c) P 1 – x 2 (P 1 – P 2 ) d) P 1 + x 2 (P 1 – P 2 )v

Solution

c) P 1 – x 2 (P 1 – P 2 )

Answer:

c) P 1 – x 2 (P 1 – P 2 )

Q.9Osomotic pressure (π) of a solution is given by the relation a) π = nRT b) πV = nRT c) πRT = n d) none of thesev

Solution

b) πV = nRT

Answer:

b) πV = nRT

Q.10Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law? a) acetone + chloroform b) water + nitric acid c) HCl + water d) ethanol + waterv

Solution

d) ethanol + water

Answer:

d) ethanol + water

Q.11The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B 0.2. The ratio of mole fraction of B and A dissolved in water will be a) \(\frac{2 x}{y}\) b) \(\frac{y}{0.2 x}\) c) \(\frac{0.2 x}{y}\) d) \(\frac{5 x}{y}\)v

Solution

d) \(\frac{5 x}{y}\)

Answer:

d) \(\frac{5 x}{y}\)

Q.12At 100°C the vapour pressure of a solution containing 6.5g a solute in 100g water is 732 mm. If K b = 0.52, the boiling point of this solution will be a) 102°C b) 100°C c) 101°C d) 100.52°Cv

Solution

c) 101°C

Answer:

c) 101°C

Q.13According to Raoults law, the relative lowering of vapour pressure for a solution is equal to a) mole fraction of solvent b) mole fraction of solute c) number of moles of solute d) number of moles of solventv

Solution

b) mole fraction of solute

Answer:

b) mole fraction of solute

Q.14At same temperature, which pair of the following solutions are isotonic? a) 0.2 M BaCl 2 and 0.2 M urea b) 0.1 M glucose and 0.2 M urea c) 0.1 M NaCl and 0.1 M K 2 SO 4 d) 0.1 M Ba(NO 3 ) 2 and 0.1 M Na 2 SO 4v

Solution

d) 0.1 M Ba(NO 3 ) 2 and 0.1 M Na 2 SO 4
Formula of normality · In the case of acid-base chemistry, normality is used to express the concentration of hydronium ions (H3O+).

Answer:

d) 0.1 M Ba(NO 3 ) 2 and 0.1 M Na 2 SO 4
Formula of normality · In the case of acid-base chemistry, normality is used to express the concentration of hydronium ions (H3O+).

Q.15The empirical formula of a non – electrolyte (X) is CH 2 O. A solution containing six grams of X exerts the same osmotic pressure as that of 0.025 M glucose solution at the same temperature. The molecular formula of X is a) C 2 H 4 O 2 b) C 8 H 16 O 8 c) C 4 H 8 O 4 d) CH 2 Ov

Solution

b) C 8 H 16 O 8

Answer:

b) C 8 H 16 O 8

Q.16The K H for the solution of oxygen dissolved in water is 4 × 10 4 atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is a) 4.6 × 10 3 b) 1.6 × 10 4 c) 1 × 10 -5 d) 1 × 10 5v

Solution

c) 1 × 10 -5

Answer:

c) 1 × 10 -5

Q.17Normality of 1.25 M sulphuric acid is a) 1.25 N b) 3.75 N c) 2.5 N d) 2.25 Nv

Solution

c) 2.5 N

Answer:

c) 2.5 N

Q.18Two liquids X and Y on mixing gives a warm solution. The solution is a) ideal b) non-ideal and shows positive deviation from Raoult’s law c) ideal and shows negative deviation from Raoult’s Law d) non-ideal and shows negative deviation from Raoult’s Lawv

Solution

d) non-ideal and shows negative deviation from Raoult’s Law

Answer:

d) non-ideal and shows negative deviation from Raoult’s Law

Q.19The relative lowering of vapour pressure of a sugar solution in water is 2.5 × 10 -3. The mole fraction of water in that solution is a) 0.0035 b) 0.35 c) 0.0035/18 d) 0.9965v

Solution

d) 0.9965

Answer:

d) 0.9965

Q.20The mass of a non – volatile solute (molar mass 80 g mol -1 ) should be dissolved in 92g of toluene to reduce its vapour pressure to 90% a) 10 g b) 20 g c) 9.2 g d) 8 gv

Solution

d) 8 g

Answer:

d) 8 g

Q.21For a solution, the plot of osmotic pressure (π) versus the concentration (c in mol L -1 ) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is a) 310 × 0.082 K b) 310° C c) 37°C d) \(\frac{310}{0.082}\) Kv

Solution

c) 37°C

Answer:

c) 37°C

Q.22200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be 2.52 × 10 -3 bar. The molar mass of protein will be (R = 0.083 L bar mol -1 K -1 } a) 62.22 kg mol -1 b) 12444 g mol -1 c) 300 g mol -1 d) None of thesev

Solution

a) 62.22 kg mol -1

Answer:

a) 62.22 kg mol -1

Q.23The Van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is a) 0 b) 1 c) 2 d) 3v

Solution

d) 3

Answer:

d) 3

Q.24Which is the molality of a 10% w/w aqueous sodium hydroxide solution? a) 2.778 b) 2.5 c) 10 d) 0.4v

Solution

b) 2.5

Answer:

b) 2.5

Q.25The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is a) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{\mathrm{n}-1}\) b) α 2 = \(\frac{n(1-i)}{(n-1)}\) c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\) d) α = \(\frac{\mathrm{n}(1-\mathrm{i})}{\mathrm{n}(1-\mathrm{i})}\)v

Solution

c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\)

Answer:

c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\)

Q.26Which of the following aqueous solutions has the highest boiling point? a) 0.1 M KNO 3 b) 0.1 M Na 3 PO 4 c) 0.1 M BaCl 2 d) 0.1 M K 2 SO 4v

Solution

b) 0.1 M Na 3 PO 4

Answer:

b) 0.1 M Na 3 PO 4

Q.27The freezing point depression constant for water is 1.86° K Kg mol -1. If 5 g Na 2 SO 4 is dissolved in 45 g water, the depression in freezing point is 3.64°C. The Vant Hoff factor for Na 2 SO 4 is a) 2.57 b) 2.63 c) 3.64 d) 5.50v

Solution

a) 2.57

Answer:

a) 2.57

Q.28Equimolal aqueous solutions of NaCl and KCl are prepared, If the freezing point of NaCl is -2°C, the freezing point of KCl solution is expected to be a) -2°C b) -4°C c) -1°C d) 0°Cv

Solution

a) -2°C

Answer:

a) -2°C

Q.29Phenol dimerizes in benzene having van’t Hoff factor 0.54. What is the degree of association? a) 0.46 b) 92 c) 46 d) 0.92v

Solution

d) 0.92

Answer:

d) 0.92

Q.30Assertion: An ideal solution obeys Raoults Law. Reason: In an ideal solution, solvent – solvent as well as solute – solute interactions are similar to solute-solvent interactions. a) both assertion and reason are true and reason is the correct explanation of assertion b) both assertion and reason are true but reason is not the correct explanation of assertion c) assertion is true but reason is false d) both assertion and reason are falsev

Solution

a) both assertion and reason are true and reason is the correct explanation of assertion
II. Write brief answer to the following questions:

Answer:

a) both assertion and reason are true and reason is the correct explanation of assertion
II. Write brief answer to the following questions:

2i. e solubility = 178. 5 × 10 52 questions

Q.46The observed depression in freezing point of water for a particular solution is 0.093°C calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K Kg mol -1.v

Solution

∆T f = 0.093°C = 0.093 K, m = ?
K f = 1.86 K Kg mol -1
∆T f = K f.m
∴ m = \(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{f}}}=\frac{0.093 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{Kg} \mathrm{mol}^{-1}}\)
= 0.05 mol Kg -1 = 0.05 m

Answer:

Q.47The vapour pressure of pure benzene (C 6 H 6 ) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?v

Solution

P°C 6 H 6 = 640 mm Hg
W 2 = 2.2 g (non volatile solute)
W 1 = 40 g (benzene)
P solution = 600 mm Hg
M 2 =?

Answer:

P°C 6 H 6 = 640 mm Hg
W 2 = 2.2 g (non volatile solute)
W 1 = 40 g (benzene)
P solution = 600 mm Hg
M 2 =?

3II. Write brief answer to the following questions:14 questions

Q.31Define: (i) Molality (ii) Normalityv

Solution

(i)Molality:
Molality (m) is defined as the number of moles of the solute dissolved in one kilogram (Kg) of the solvent. The units of molality are moles per kilogram, i.e., mole kg -1. The molality is preferred over molarity if volume of the solution is either expanding or contracting with temperature.
molality (m) = \(\frac{\text { Number of mole of solute }}{\text { mess of solvent in } \mathrm{kg}}\)
ii) Normality:
Normality (N) of a solution is defined as the number of gram equivalents of the solute present in one liter of the solution. Normality is used in acid-based redox titrations.
Normality (N) = \(\frac{\text { Number of gram equivalents of solute }}{\text { Volume of solution in litre }}\)

Answer:

Q.32a) What is a vapour pressure of liquid?v

Solution

“The pressure exerted by the vapors above the liquid surface which is in equilibrium with the liquid at a given temperature is called vapor pressure”.
b) What is a relative lowering of vapour pressure?
The relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to the vapour pressure
of pure solvent (P 0 ) RLVP = \(\frac{p^{0}-P}{P^{0}}\)

Answer:

Q.33State and explain Henry’s law.v

Solution

Henry’s law:
This law states “that the partial pressure of the gas in vapour phase is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations.”
P solute ∝ x solute in solution
P solute = K H. x solute in solution
x solute = mole fraction of solute in the solution
K H = empirical constant.
P solute = Vapour pressure of the solute (or) the partial pressure of the gas in vapour state. The value of K H depends on the nature of the gaseous solute and solvent.

Answer:

Q.34State Raoult law and obtain the expression for lowering of vapour pressure when the nonvolatile solute is dissolved Insolvent.v

Solution

In an ideal solution, the vapour pressure of the solution is decreased when a non-volatile solute is dissolved in a solvent. The magnitude of decrease in the vapour pressure of the solution depends on the amount of solute added.
Let us consider the solution with the following features.
Mole fraction of the solvent = x A
Mole fraction of the solute = x B
Vapour pressure of the pure solvent = P° A
Vapour pressure of solution = P
As the solute is nonvolatile, the vapour pressure of the solution is only due to the solvent. Therefore, the vapour pressure of the solution (P) will be equal to the vapour pressure of the solvent (P A ) over the solution.
i.e., P = P A
According to Raoult’s law, the vapour pressure of solvent over the solution is equal to the product or its vapour pressure in a pure state and its mole fraction.
P A = P° A x A or
P = P° A x A

Answer:

Q.35What is molal depression constant? Does it depend on nature of the solute?v

Solution

If m = 1 then ∆T f = K f
“Then K f is equal to the depression in freezing point for 1 molal solution”. No, it does not depends on the nature of the solute.

Answer:

If m = 1 then ∆T f = K f
“Then K f is equal to the depression in freezing point for 1 molal solution”. No, it does not depends on the nature of the solute.

Q.36What is osmosis?v

Solution

Osmosis is a spontaneous process by which the solvent molecules pass through a semipermeable membrane from a solution of lower concentration to the solution of higher concentration.

Answer:

Osmosis is a spontaneous process by which the solvent molecules pass through a semipermeable membrane from a solution of lower concentration to the solution of higher concentration.

Q.37Define the term ‘isotonic solution’.v

Solution

Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

Answer:

Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

Q.38You are provided with a solid. ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one supersaturated. How would you determine which solution is which?v

Solution

(A) Unsaturated solution:
It can dissolve salt in addition to it.
(B) Saturated solution:
Further solubility of salt does not take place but solubility can take place on heating.
(c) Supersaturated solution:
Solubility of salt does not take place on even further heating.

Answer:

Q.39Explain the effect of pressure on solubility.v

Solution

1. The change in pressure does not have any significant effect on the solubility of solids and liquids as they are not compressible. However, the solubility of gases generally increases with the increase in pressure.
2. According to Le – chatter’s principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more gaseous molecules dissolve in the solvent.
3. If pressure increases, the solubility of gas also increases.

Answer:

Q.40A sample of 12 M Concentrated hydrochloric acid has a density 1.2 M gL -1 calculate the molality.v

Solution

Given:
Molarity = 12 M HCl
density of solution = 1.2 g L -1
In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution.
Molality = \(\frac{\text { no of moles of solute }}{\text { mass of solvent (in } \mathrm{kg} \text { ) }}\)
Calculate mass of water(solvent)
mass of 1 litre HCl solution = density × volume
= 1.2 × gmL -1 × 1000 mL = 1200 g
mass of HCl = no. of moles of HCl × molar mass of HCl
= 12 mol × 36.5 g mol -1
= 438 g.
mass of water = mass of HCl solution – mass of HCl
mass of water = 1200 – 438 = 762 g
molality(m) = \(\frac{12}{0.762}\) = 15.75 m

Answer:

Q.41A 0.25 M glucose solution, at 370.28 K has approximately the pressure as blood does what is the osmotic pressure of blood?v

Solution

C = 0.25 M
T = 37O.28 K
(π) gIucose = CRT
(π) = 0.25 mol L -1 x 0.082 L atm K -1 morl -1 x 370.28K
= 7.59 atm

Answer:

C = 0.25 M
T = 37O.28 K
(π) gIucose = CRT
(π) = 0.25 mol L -1 x 0.082 L atm K -1 morl -1 x 370.28K
= 7.59 atm

Q.43Which solution has the lower freezing point? 10 g of methanol (CH 3 OH) in 100g g of water (or) 20 g of ethanol (C 2 H 5 OH) in 200 g of water.v

Solution

∆T f = K f i.e
∆T f α m
m CH 3 -OH = \(\frac{\left(\frac{10}{32}\right)}{0.1}\)
= 3.125 m
m C 2 H 5 -OH = \(\frac{\left(\frac{20}{46}\right)}{0.2}\)
= 2.174 m
∴ Depression in freezing point is more in methanol solution and it will have lower freezing point.

Answer:

Q.44How many moles of solute particles are present in one liter of 10 -4 M potassium sulphate?v

Solution

In 10 -4 M K 2 SO 4 solution, there are 10 -4 moles of potassium sulphate.
K 2 SO 4 molecule contains 3 ions (2K+ and 1 SO 4 2- )
1 mole of K 2 SO 4 molecule contains 3 × 6.023 × 10 23 ions
10 -4 mole of K 2 SO 4 contains 3 × 6.023 × 10 23 × 10 -4 ions
= 18. 069 × 10 19

Answer:

Q.45Henry’s law constant for solubility of methane in benzene is 4.2 × 10 -5 mm Hg at a particular constant temperature. At this temperature calculate the solubility of methane at i) 750 mm Hg ii) 840 mm Hg.v

Solution

(K H ) benzene = 4.2 × 10 -5 mm
Solubility of methane =?
P = 750 mm Hg P = 840 mm Hg
According to Henrys Law,
P = K H X in solution
750 mm Hg = 4.2 × 10 -5 mm Hg. X in solution
⇒ X in solution = \(\frac{750}{4.2 \times 10^{-5}}\)
i. e solubility = 178. 5 × 10 5
similarly at P = 840 mm Hg
solubility = \(\frac{840}{4.2 \times 10^{-5}}\) = 200 × 10 -5

Answer:

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