## Topics in CBSE Solutions for Class 12 Chemistry Chapter 2 Solutions:

Topic Covered |
---|

Solutions |

Types of Solutions |

Expressing Concentration of Solutions |

Solubility |

Vapour Pressure of Liquid Solutions |

Ideal and Non-ideal Solutions |

Colligative Properties and Determination of Molar Mass |

Abnormal Molar Masses |

## CBSE or NCERT Textbook Question Solved

### 2.1 Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride

**Ans: **Mass % of benzene

Mass percentage of C6H6=(Mass of C6H6)/(Mass of solution)×100 = 22×100/22+122=15.28%

Mass percentage of ℂl4 =(Mass of CCL4)/(Mass of solution)×100 = 122×100/22+122=84.72%

### 2.2 Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride

**Ans:** Let the total mass of the solution be 100 g and the mass of benzene be 30 g.

∴ Mass of carbon tetrachloride = (100 – 30) g = 70 g

Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol – 1= 78 g mol – 1

∴ Number of moles of C6H6 =30/78 mol= 0.3846 mol

Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 35.5= 154 g mol – 1

∴ Number of moles of CCl4 = 70/154 mol= 0.4545 mol

Thus, the mole fraction of C6H6 is given as:

= 0.3846 / (0.3846 + 0.4545)= 0.458

### 2.3 Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution

(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

**Ans:** Molarity is given by

**(A)** Given : Co(NO3 )2. 6H2O = 30 gm

Volume of solution = 4.3 liter

We know :

Molar mass of Co(NO3 )2. 6H2O = 59 + 2(14 + 3 × 16) + 6 × 18 = 291 g mol−1

Moles of (NO3 )2. 6H2O = (weight)(gm)

molecularweight = 30 291 mol = 0.103 mol

The formula used : Molarity = Moles of solute Volume of solution in litre

Therefore, molarity= 0.103 mol 4.3 L = 0.023 M

**(B)** Given : 30 mL of 0.5 M H2SO4

Initial Volume of H2SO4 solution =30 ml

Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol

Number of moles present in 30 mL of 0.5 M H2SO4

= 0.5×30 1000 mol = 0.015 mol

Therefore,Molarity = Moles of solute/ Volume of solution in litre

= Moles of solute×1000 /Volume of solution in Mili litre

Moles of solute = MV(litre)

Final Volume of the solution after dilution = molarity = MiVi/ Vf

= 0.015/0.5 L mol=0.03 M

### 2.4 Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution. 2.5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1

**Ans:** Molar mass of urea (NH2CONH2)= 2 (1×14+2×1)+1×12+1×16= 60g mol – 1

0.25 molar aqueous solution of urea means:

1000 g of water contains 0.25 mol= (0.25 × 60) g of urea = 15 g of urea

That is, (1000 + 15) g of solution contains 15 g of urea

Therefore, 2.5 kg (2500 g) of solution contains = (15 X 2500) / (1000+15) g= 36.95 g= 37 g of urea (approximately)

Hence, mass of urea required= 37 g

### 2.5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1

**Ans:** **(A)** Molar mass of KI = 39 + 127 = 166 g mol – 1

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.

That is,

20 g of KI is present in (100 – 20) g of water = 80 g of water

Therefore, molality of the solution = Moles of KI / Mass of water in kg

= 20/166 / 0.08 m

= 1.506 m

= 1.51 m (approximately)

**(B)** It is given that the density of the solution = 1.202 g mL – 1

∴Volume of 100 g solution = Mass / Density

= 100g / 1.202g mL-1

= 83.19 mL

= 83.19 × 10 – 3 L

Therefore, molarity of the solution = 20/166 mol / 83.19 × 10 – 3 L= 1.45 M

**(C)** Moles of KI = 20/166 = 0.12 mol

Moles of water = 80/18 = 4.44 mol

Therefore, mole fraction of KI = Moles of KI / (Moles of KI + Moles of water)

= 0.12 / (0.12+4.44)= 0.0263

### 2.6 H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant

**Ans:** 1kg of solvent contains 181000=55.55 moles.

Mole fraction of H2S is 0.195+55.550.195=55.4750.195=0.0035

Pressure of H2S =0.03550.987=282bar

( at STP pressure is 0.987bar.)

### 2.7 Henry’s law constant for CO2 in water is 1.67×108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.

**Ans: **According to Henry’s law, pCO2 = KHx

Here x is the mole fraction of carbon dioxide in the carbon dioxide water mixture.

pCO2 = 2.5 atm = 2.5 * 1.01325 * 105 Pa = 2.533125 * 105 Pa

x = mole fraction = 2.533125 * 105/ 1.67 * 108 = 0.00152

x = moles of CO2/(moles of CO2 + moles of H2O) ≈moles of CO2/moles of H2Oas mole of CO2are negligible in comparison to the number of moles of water

500 ml of water = 500g of water = 500/18 = 27.78 moles of water.

So, moles of CO2/27.78 = 0.00152 or moles of CO2 = 0.042 moles.

Hence the amount of CO2 in 500 ml of soda water is 0.042 * 44 = 1.848 g

### 2.8 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

**Ans:** Let the composition of liquid A (mole fraction) be xA.

So mole fraction of B will be xB = 1 – xA.

Given that,

Using Raoult’s law ,

Putting values of ptotal and vapour pressure of pure liquids in the above equation, we get :

600 = 450.xA + 700.(1 – xA)

or 600 – 700 = 450xA – 700xA

or xA = 0.4

and xB = 0.6

Now pressure in vapour phase :

= 450(0.4) = 180 mm of Hg

= 700(0.6) = 420 mm of Hg And mole fraction of liquid B = 0.70

### 2.9 Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2 ) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

**Ans: **Given, Vapour pressure of water, PIo = 23.8 mm of Hg

Weight of water, w1 = 850 g

Weight of urea, w2 = 50 g

Molecular weight of water, M1 = 18 g/mol

Molecular weight of urea, M2 = 60 g/mol

n_{1}=w_{1}/M_{1}=850/18=47.22 mol

n_{2}=w_{2}/M_{2}=50/60=0.83 mol

We have to calculate vapour pressure of water in the solution p_{1}

By using Rault’s theorem,

P_{1} = 23.4 mm of Hg

Hence,

The vapour pressure of water in the solution is 23.4mm of Hg and its relative lowering is 0.0173.

### 2.10 Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

**Ans:** Mass of water, **w _{1}** = 500 g

Boiling point of water = 99.63°C (at 750 mm Hg).

Molal elevation constant,

**K**= 0.52 K kg/mol

_{b}Molar mass of sucrose (C12H22O11),

**M**=> (11 × 12 + 22 × 1 + 11 × 16) = 342 g/mol

_{2}Elevation of boiling point

**ΔT**= (100 + 273) – (99.63 + 273) = 0.37 K

_{b}We know that,

**ΔTb** = K_{b}*1000*w_{2} / M_{2}*w_{1}

0.37 = 0.52*1000*w_{2} / 342*500

**w _{2}** = 0.37*342*500 / 0.52*1000

**w _{2}** = 121.67 g

Hence, 121.67 g (approx) Sucrose is added to 500g of water so that it boils at 100°C.

### 2.11 Calculate the mass of ascorbic acid (Vitamin C, C6H8O6 ) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol-1 .

**Ans:** Mass of acetic acid, w_{1}= 75 g Molar mass of ascorbic acid (C6H8O6),

M_{2}=6×12+8×1+6×16 =176gmol−1

Lowering of melting point, **ΔTf**=1.5K

We know that:**ΔTb** =M2×w1Kb×1000×w2

w_{2 }=Kb×1000ΔTb×M2×w1

w_{2} =3.9×10001.5×176×75

w_{2} =5.08g

Hence, 5.08 g of ascorbic acid is needed to be dissolved.

### 2.12 Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

**Ans:** Osmotic Pressure= π =(n/v)RT

We are given with: Moles of Polymer=1/185000

Volume, V = 0.45 L

Thus osmotic pressure= [ (1/185000) * 8.314 * 10^{3} * 310 ] / 0.45 = 30.98Pa