Class 12 Chemistry

# Solutions

## NCERT Solution Part 2

Question 11: Why do gases always tend to be less soluble in liquids as the temperature is raised?

**Answer:** Dissolution of a gas in a liquid is an exothermic process. Hence, gases show less solubility in liquid when the temperature is increased. It happens in order to obey Le Chatelier’s Principle.

Question 12: State Henry’s law and mention some important applications.

**Answer:** Henry’s Law: At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution. Henry’s Law is also stated as, “the partial pressure of the gas in vapor phase (p) is proportional to the mole fraction of the gas (x) in the solution. It is expressed as follows:

`p=K_H\X` …………….. (2)

Here, `K_H` is the Henry’s law constant.

Following are some applications of Henry’s Law:

- While sealing the bottle of soft drinks
- Packing oxygen cylinders for scuba divers
- Breathing difficulty at high altitude

Question 13: The partial pressure of ethane over a solution containing 6.56 `xx` 10^{-2} g of ethane is 1 bar. If the solution contains 5.00 `xx` 10^{-2} g of ethane, then what shall be the partial pressure of the gas?

**Answer:** We know that `m=K_H\xx\P`

So, in first condition:

`6.56xx10^(-2)=K_H\xx1` bar

Similarly, in second condition:

`5.00xx10^(-2)=K_H\xx\P`

From both equations, we have

`K_H=(6.56xx10^(-2))/1=(5.00xx10^(-2))/P`

Or, `P=(5.00xx10^(-2))/(6.56xx10^(-2))=0.762` bar

Question 14: What is meant by positive and negative deviations from Raoult’s law and how is the sign of Δ_{mix}H related to the positive and negative deviations from Raoult’s law?

**Answer:** Let us assume that molecules A and B make a solution. Positive deviation from Raoult’s law means A-B interaction is weaker than A-A and B-B interactions. A negative deviation means A-B interaction is stronger than A-A and B-B interactions. In case of positive deviation the enthalpy of mixing is positive, i.e. Δ_{mix}H > 0. It means that volume of solution is greater than sum of volumes of solute and solvent. In case of negative deviation the enthalpy of mixing is negative, i.e. Δ_{mix}H < 0. It means that volume of solution is less than sum of volumes of solute and solvent.

Question 15: An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

**Answer:** Let us assume that mass of solution is 100 g

Then mass of solute = 2 g and mass of solvent (water) = 98 g

We know, vapor pressure of pure water at boiling point = 1.013 bar

Vapor pressure of solution = 1.004 bar

`(P_o-P_s)/(P_s)=(n_2)/(n_1)`

Or, `(P_o-P_s)/(P_s)=((W_2)/(M_2))/((W_1)/(M_1))`

Or, `(P_o-P_s)/(P_s)=(W_2xxM_1)/(W_1xxM_2)`

Or, `(1.013-1.004)/(1.013)=(2xx18)/(M_2xx98)`

Or, `(0.009)/(1.013)=(18)/(M_2xx49)`

Or, `M_2=(18xx1.013)/(49xx0.009)=41.35` g mol^{-1}

Question 16: Heptane and octane form an ideal solution. At 373 K, the vapor pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapor pressure of a mixture of 26.0 g of heptanes and 35 g of octane?

**Answer:** Molar mass of heptane (C_{7}H_{16})

`=7xx12+16=84+16=100` g mol^{-1}

Molar mass of octane (C_{8}H_{18})

`=8xx12+18=114` g mol^{-1}

Number of moles of heptane `=(26.0)/(100)=0.26` mole

Number of moles of octane `=(35.0)/(114)=0.307` mole

Mole fraction of heptane `=(0.26)/(0.26+0.307)=(0.26)/(0.567)=0.458`

So, mole fraction of octane `=1-0.458=0.542`

Vapr pressure of heptane `=X_H\xx\P_O`

`=0.458xx105.2 kPa=48.18 kPa`

Vapor pressure of octane `=X_O\xx\P_O`

`=0.542xx46.8 kPa=25.36 kPa`

So, vapor pressure of mixture `=48.18+25.36=73.54 kPa`

Question 17: The vapor pressure of water is 12.3 kPa at 300 K. Calculate vapor pressure of 1 molar solution of a non-volatile solute in it.

**Answer:** 1 Molal solution means 1 mole of solute is present in 1000 g of solvent

Molar mass of solvent (water) = 18 g mol^{-1}

Mole fraction of solute `=1/(1+55.5)=0.018`

`(P_o-P_s)/(P_o)=X_2`

Or, `(12.3-P_s)/(12.3)=0.018`

Or, `12.3-P_s=0.018xx12.3=0.21744`

Or, `P_s=12.3-0.21744=12.08` kPa

This is the vapor pressure of given solution

Question 18: Calculate the mass of a non-volatile solute (molar mass 40 g mol^{-1}) which should be dissolved in 114 g octane to reduce its vapor pressure to 80%.

**Answer:** Given `P_s=0.8P_o`

Let us assume that mass of solute = W gram

So, moles of solute `=W/(40)`

Molar mass of octane= 114 g mol^{-1}

Number of moles of octane `=(114)/(114)=1` mole

Mole fraction of solute `=(W/(40))/(W/(40)+1)`

`=(W/(40))/((W+40)/(40))`

`=W/(W+40)`

We know that `(P_o-P_s)/(P_o)=x_2`

Or, `(P_o-0.8P_o)/(P_o)=W/(W+40)`

Or, `(0.2P_o)/(P_o)=W/(W+40)`

Or, `W=0.2(W+40)=0.2W+8`

Or, `W-0.2W=8`

Or, `0.8W=8`

Or, `W=8/(0.8)=10` g

This is the required mass to reduce vapor pressure to desired level.

Question 19: A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapor pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapor pressure becomes 2.9 kPa at 298 K. Calculate (i) molar mass of the solute (ii) vapor pressure of water at 298 K.

**Answer:** Moles of solvent (water) `=(90)/(18)=5` moles

Moles of solute `=(30)/M`

`(P_o-P_s)/(P_o)=(n_2)/(n_1+n_2)`

Or, `(P_o-2.8)/(P_o)=((30)/M)/(5+(30)/M)`

Or, `1-(2.8)/(P_o)=(30)/(5M+30)`

Or, `1-(30)/(5M+30)=(2.8)/(P_o)`

Or, `1-6/(M+6)=(2.8)/(P_o)`

Or, `(M+6-6)/(M+6)=(2.8)/(P_o)`

Or, `M/(M+6)=(2.8)/(P_o)`

Or, `(P_o)/(2.8)=1+6/M` ………..(1)

After adding 18 gram of water

Moles of water becomes 6 moles

Then, `(P_o-P_s)/(P_o)=(30)/(6M+30)`

Or, `(P_o-2.9)/(P_o)=5/(M+5)`

Or, `1-(2.9)/(P_o)=5/(M+5)`

Or, `1-5/(M+5)=(2.9)/(P_o)`

Or, `(M+5-5)/(M+5)=(2.9)/(P_o)`

Or, `M/(M+5)=(2.9)/(P_o)`

Or, `(P_o)/(2.9)=1+5/M` ………….(2)

Dividing equation (1) by (2)

`(2.9)/(2.8)=(M+6)/MxxM/(M+5)=(M+6)/(M+5)`

Or, `2.9M+14.5=2.8M+16.8`

Or, `2.9M-2.8M=16.8-14.5`

Or, `0.1M=2.3`

Or, `M=(2.3)/(0.1)=23` g mol^{-1}

Putting M in equation (1)

`(P_o)/(2.8)=1+6/(23)`

Or, `P_o=2.8xx(29)/(23)=3.53` kPa

Question 20: A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose water if freezing point of pure water is 273.15 K.

**Answer:** If solvent (water) is 100 g then sugar is 5%, i.e. 5 g

Molar mass of sugar = 342 g mol^{-1}

Hence, molality of sugar `=(5xx1000)/(342xx100)=0.146`

So, ΔT_{f} for sugar solution

`=273.15-271=2.15 K`

`ΔT_f=K_f\xx\m`

Or, `2.15=K_f\xx0.146`

Or, `K_f=(2.15)/(0.146)`

Now, molar mass of glucose = 180 g mol^{-1}

Hence, molality of glucose `=5/(180)xx(1000)/(100)=0.278`

Here, `ΔT_f=K_f\xx\m`

`=(2.15)/(0.146)xx0.278=4.09` K

So, freezing point of glucose solution

`=273.15-4.09=269.06` K

Molality

InText Questions

NCERT Solution 1

NCERT Solution 2

NCERT Solution 3

NCERT Solution 4