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Chapter: 11th Chemistry : UNIT 9 : Solutions

Answer the following questions: Chemistry: Solutions

Chemistry: Solutions - Brief questions and answers

Chemistry: Solutions

Answer the following questions

31. Define (i) molality   (ii) Normality

(i) Molality:

Molality of the solution is defined as number of gram moles of the solute dissolved in 1kg of the solution

Molality = Number of moles of solute / Mass of the solvent (in kg)

(ii) Normality:

Normality of the solution is defined as number of gram equivalent of the solute dissolved in one litre of the solution

Normality = Number of gram equivalent of solute / Volume of solution (in L) 


32. What is a vapour pressure of liquid? What is relative lowering of vapour pressure?

i) Vapour pressure of Liquid :

The vapour pressure of a liquid is the pressure exerted by its vapour, pressure when it is in dynamic equilibrium with its liquid, in closed container.

ii) Relative lowering of vapour pressure :

● On addition of a non-volatile solute it is observed that the vapour pressure of the solution is less than the solvent.

According to Raolut's law : The relative lowering of vapour pressure, is defined as the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent. 


33. State and explain Henry’s law

● The partial pressure of gas in vapour phase (vapour pressure of solute) is directly proportional to the Mole fraction (x) of the gaseous solute in the solution at low concentrations.

● This statement is called Henry's Law.

Psolute Xsolute in solution

Psolute = KH Xsolute in solution

Psolute → Partial pressure of solute

Xsolute in solution → mole fraction of solute in the solution

KH → a empirical constant with the dimensions of pressure.


34. State Raoult law and obtain expression for lowering of vapour pressure when nonvolatile solute is dissolved in solvent.

Raoult's Law:

The vapour pressure of a solution containing a non-volatile solute is directly proportional to the mole fraction of the solvent.

Expression for relative lowering of vapour pressure:

● The measurement of relative lowering of vapour pressure can be used to determine the molar mass of non-volatile solute.

● A known mass of the solute is dissolved in a know quantity of solvent.

● The relative lowering of vapour pressure is measured experimentally.

● According to Raoult's Law, the relative lowering of vapour pressure is,

[ PoSolvent − PoSolvent ] / PoSolvent = XB         ……………… 1

Let WA and WB be the weight of solvent and solute respectively and their corresponding molar masses are MA and MB then its mole fraction solute is XB. It is given by

XB = nB / [nA + nB] ………………. 2

Where nA and nB are the moles of the solvent and solute respectively.

For dilute solutions, nA >> nB Hence nA + nB ≈ nA

XB = nB / nA ………………… 3

We know that

nB = WB / MB …………….. 4

nA = WA / MA …………….. 5

XB = (WB/MB) / (WA / MA) …………6

The relative lowering of vapour Pressure = [ WB / MB ] × [ MA / WA ] ………….(7)

From the equation (1) molar mass of the solute can be calculated, If the value of WA, WB, MA and measured values of relative vapour pressure are known.


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

● Molal freezing point depression constant or cryoscopic constant (Kf) is defined as the depression in freezing point produced when mole of solute is dissolved in 1kg solvent. 

● Freezing point depression of dilute solution is found to be directly proportional to the number of moles of the solute dissolved in a given amount of solvent.

● ΔTf is independent of the nature of the solute as long as it is non-volatile.


36. What is osmosis?

Spontaneous movement of solvent particles from a solution of lower concentration to a solution of higher concentration through a semipermeable membrane is known as osmosis.


37. Define the term ‘isotonic solution’.

Two solutions of different substance having same osmotic pressure at given temperature are called isotonic solutions.


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

If more solute (A) is added and it does not dissolve, then the solution was saturated solution.

● If the added solute (A) dissolves, then the solution was unsaturated solution.

● If the added solute (A) does not dissolves but it settle on the bottom of vessel is called super saturated solution. In this way we can determine which solution is which.

39. Explain the effect of pressure on the solubility.

● In general, the change in pressure does not have any significant effect in the solubility of solids and liquids because they are non-compressible.

● Solubility of gases generally increase with increase of pressure.

● A saturated solution of gaseous solute dissolved in a liquid solvent in a closed container.

Gas (in gaseous state) gas (in solution)

● According to Le-chateliers principle, the increase in pressure will shift the equilibrium in the direction which will reduced the pressure.

● Therefore more number of gaseous molecules dissolves in the solution and the solubility increases. 


40. A sample of 12 M Concentrated hydrochloric acid has a density 1.2 gL–1 Calculate the molality


Molality = 12 M HCl

density of the solution = 1.2 gL−1

In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution

Molality = no. of moles of solute / mass of solvent (in Kg)

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 = 12 / 0.762

= 15.75 m


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


C = 0.25m

T= 370.28 K

(π)glucose = CRT

(π) = 0.25molL−1 × 0.082L atm K−1 mol−1 × 370.28K

= 7.59 atm


42. Calculate the molality of a solution containing 7.5 g of glycine (NH2-CH2 -COOH) dissolved in 500 g of water.


molality = no. of moles of solute / mass of solvent (in Kg)

no. of moles of glycine = mass of glycine / molar mass of glycine = 7.5 / 75 = 0.1

molality = 0.1 / 0.5 Kg = 0.2 m 


43. Which solution has the lower freening point?10 g of methanol (CH3OH) in 100g of water (or) 20 g of ethanol (C2H5OH) in 200 g of water.


ΔTf = Kf m ; ie ΔTf m

m(CH3OH) = (10/32) / 0.1 = 3.125m

M(C2H5OH)  = (20/46) / 0.2 = 2.174m

depression in freezing point is more in methanol solution and it will have lower freezing point.


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


In 10-4 M K2SO4 solution, there are 10-4 moles of potassium sulphate.

K2SO4 molecule contains 3 ions (2K+ and 1 SO42− )

1 mole of K2SO4 contains 3 × 6.023×1023 ions

10-4 mole of K2SO4 contains 3 × 6.023×1023 × 10−4 ions

= 18.069×1019


45. Henry’s law constant for solubility of methane in benzene is 4.2x10-5 mm Hg at a particular constant temperature At this temperature. Calculate the solubility of methane at i) 750 mm Hg ii) 840 mm Hg


(kH )benzene = 4.2 × 10-5 mm Hg ;

Solubility of methane = ?

P = 750mm Hg

P = 840 mm Hg

According to Henry's Law

P = KH.x in solution

750mm Hg = 4.2 x 10-5 mm Hg.xin solution

xin solution = 750 / (4.2×10-5)

i.e. solubility, = 178.5 × 105 = 1.785 × 105

similarly at P = 840 mm Hg

solubility = 840 / 4.2×10-5

 = 200 × 10-5


46. The observed depression in freezing point of water for a particular solution is 0.093ºC. Calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 KKg mol-1


ΔTf = 0.093°C = 0.093K ;

m = ?

Kf = 1.86 K Kg mol−1 ;

ΔTf = Kf.m

μ = ΔTf / Kf

m = 0.093K / 1.86 K Kg mol−1

= 0.05 mol Kg −1 = 0.05 m. 


47. The vapour pressure of pure benzene (C6H6) 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?


C6H6 = 640 mm Hg

W2 = 2.2 g (non volatile solute)

W1 = 40 g (benzene)

Psolution = 600mm Hg

M2 = ?

(P°−P) / P° = X2 ;

(640 – 600) / 640 = n2 / n1 + n2       [n1 >> n2 ; n1 + n2 = n1 ]

40 / 640 = n2 / n1

0.0625 = ( W2 × M1 ) / (M2 × W1)

M2 = (2.2 × 78) / (0.0625 × 40) = 68.64 g mol−1

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