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Brief questions and answers: Chemistry: Solutions

Chemistry: Solutions - Brief questions and answers

31. Define (i) molality   (ii) Normality

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

33. State and explain Henry’s law

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

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

36. What is osmosis?

37. Define the term ‘isotonic solution’.

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 ?  39. Explain the effect of pressure on the solubility.

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

Solution:

Given

Molarity = 12 M HCl

density of the solution = 1.2 g L–1

In 12M 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 g mL–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 = 15.75m

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 ?

Solution:

C = 0.25 M

T = 370.28 K

(π)glucose = CRT

(π)=0.25 mol L–1 × 0.082L atm K–1mol–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.

Solution:

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

no. of moles of glycine = mass of glycine / molar mass of glycinee

= 7.5/75 = 01

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.

Solution:

∆Tf = Kf. m

ie ∆Tf α m 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?

Solution:

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

K2SO4 molecule contains 3 ions (2K+ and 1So42–)

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

Solution:

(kH)bonzene = 4.2 × 10–5 mm Hg

Solubility of methane = ?

P = 750mm Hg

P = 840 mm Hg

According to Henrys Law,

P = KH . xin solution.

750mm Hg = 4.2 × 10–5 mm Hg . xin solution

­xinsolution = 750/4.2x10-5.

i.e, solubility = 178.5 × 105

similarly at P = 840 mm Hg

solubility = 840/4.2x10-5

=200x10-5.

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

Solution:

∆Tf = 0.093ºC = 0.093K

m = ?

Kf = 1.86K Kg mol–1

∆Tf = Kf . m

m = ∆Tf / Kf

= [ 0.093K ] / [1.86 K Kgmol-1]

= 0.05 mol Kg-1

= 0.05m

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?

Solution:

C6H6 = 640 mm Hg

W2 = 2.2 g (non volabile solute)

W1 = 40 g (benzene)

Psolution = 600 mm Hg

M2 = ? Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

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