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**Q.****15. Define the following terms:****(a) Mole fraction****(b) Molality****(c) Molarity ****(d) Mass percentage.****Ans.****(a) Mole fraction**

The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture.

i.e.,

Mole fraction of a component

Mole fraction is denoted by ‘*x*’.

If in a binary solution, the number of moles of the solute and the solvent are *n _{A}* and

by,

Similarly, the mole fraction of the solvent in the solution is given as:

**(b) Molality**

Molality (m) is defined as the number of moles of the solute per kilogram of the

solvent. It is expressed as:

Molality (m)

**(c) Molarity**

Molarity (M) is defined as the number of moles of the solute dissolved in one Litre of the solution.

It is expressed as:

Molarity (M)

**(d) Mass percentage**

The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the solution. It is expressed as:

Mass % of a component**Q.****16. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL ^{−1}?**

Molar mass of nitric acid (HNO

Then, number of moles of HNO

Given,

Density of solution = 1.504 g mL

Volume of 100 g solution = 100 / 1.504 mL

= 66.49 mL

= 66.49 × 10

Molarity of solution

= 16.23 M

Molar mass of glucose (C

Then, number of moles of glucose = 10/180 mol = 0.056 mol

Molality of solution = 0.056 mol / 0.09 kg = 0.62 m

Number of moles of water = 90g / 18g mol

Mole fraction of glucose (x

= 0.011

And, mole fraction of water

= 1 − 0.011 = 0.989

If the density of the solution is 1.2 g mL

∴ Molarity of the solution

= 0.67 M

Let the amount of Na_{2}CO_{3} in the mixture be *x* g.

Then, the amount of NaHCO_{3} in the mixture is (1 − *x*) g.

Molar mass of Na_{2}CO_{3} = 2 × 23 + 1 × 12 + 3 × 16 = 106 g mol^{−1}

Number of moles Na_{2}CO_{3 }= *x* / 106 mol

Molar mass of NaHCO_{3} = 1 × 23 + 1 × 12 + 3 × 16 = 84 g mol^{−1}

Number of moles of NaHCO_{3} = 1 -x /84 mol

According to the question,

⇒ 84*x* = 106 − 106*x*

⇒ 190*x* = 106

⇒ *x* = 0.5579

Therefore, number of moles of Na_{2}CO_{3}

= 0.0053 mol

And, number of moles of NaHCO_{3}

= 0.0053 mol

HCl reacts with Na_{2}CO_{3} and NaHCO_{3} according to the following equation.

1 mol of Na_{2}CO_{3} reacts with 2 mol of HCl.

Therefore, 0.0053 mol of Na_{2}CO_{3} reacts with 2 × 0.0053 mol = 0.0106 mol.

Similarly, 1 mol of NaHCO_{3} reacts with 1 mol of HCl.

Therefore, 0.0053 mol of NaHCO_{3} reacts with 0.0053 mol of HCl.

Total moles of HCl required = (0.0106 + 0.0053) mol

= 0.0159 mol

In 0.1 M of HCl,

0.1 mol of HCl is present in 1000 mL of the solution.

Therefore, 0.0159 mol of HCl is present in

= 159 mL of the solution

Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na_{2}CO_{3} and NaHCO_{3,} containing equimolar amounts of both.**Q.****19. 100 g of liquid A (molar mass 140 g mol ^{−1}) was dissolved in 1000 g of liquid B (molar mass 180 g mol^{−1}). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.**

= 0.714 mol

Number of moles of liquid B,

= 5.556 mol

Then, mole fraction of A,

= 0.114

And, mole fraction of B, *x*_{B} = 1 − 0.114

= 0.886

Vapour pressure of pure liquid B, p_{b}^{o} = 500 torr

Therefore, vapour pressure of liquid B in the solution,

p_{b} = p_{b}^{o}x_{b}

= 500 × 0.886

= 443 torr

Total vapour pressure of the solution, *p*_{total} = 475 torr

Vapour pressure of liquid A in the solution,*p*_{A} = *p*_{total} − *p*_{B}

= 475 − 443

= 32 torr

Now,

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.**Q.****20. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C _{2}H_{6}O_{2}) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^{−1}, then what shall be the molarity of the solution?**

Number of moles of ethylene glycol

= 3.59 mol

Therefore, molality of the solution = 3.59 MOL / 0.200 kg = 17.95 m

Total mass of the solution = (222.6 + 200) g = 422.6 g

Given,

Density of the solution = 1.072 g mL

Volume of the solution = 422.6 g / 1.072g mL

= 394.22 mL

= 0.3942 × 10

⇒ Molarity of the solution

= 9.11 M

Therefore, percent by mass

= 1.5 × 10

= 119.5 g mol

Now, according to the question,

15 g of chloroform is present in 10

i.e., 15 g of chloroform is present in (10

Molality of the solution

= 1.26 × 10

Gas + Liquid → Solution + Heat

Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.

Where,

Some important applications of Henry’s law are mentioned below.

Hence, the oxygen tanks used by scuba divers are filled with air and diluted with helium to avoid bends.

**Q.****25. ****T****he partial pressure of ethane over a solution containing 6.56 × 10 ^{−3} g of ethane is 1 bar. If the solution contains 5.00 × 10^{−2} g of ethane, then what shall be the partial pressure of the gas?**

Molar mass of ethane (C

Number of moles present in 6.56 × 10

= 2.187 × 10

Let the number of moles of the solvent be

According to Henry’s law,

Number of moles present in 5.00 × 10

= 1.67 × 10

According to Henry’s law,

= 7.636 bar

Hence, partial pressure of the gas shall be 7.636 bar.

In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.

Δ

In the case of solutions showing positive deviations, absorption of heat takes place.

∴ Δ

In the case of solutions showing negative deviations, evolution of heat takes place.

∴ Δ

Vapour pressure of the solution at normal boiling point (

Vapour pressure of pure water at normal boiling point (

Mass of solute, (

Mass of solvent (water), (

Molar mass of solvent (water), (

According to Raoult’s law,

= 41.35 g mol

Hence, the molar mass of the solute is 41.35 g mol

Vapour pressure of octane (p°

We know that,

Molar mass of heptane (C

= 100 g mol

Number of moles of heptane = 26 / 100 mol

= 0.26 mol

Molar mass of octane (C

= 114 g mol

Number of moles of octane = 35/114 mol

= 0.31 mol

Mole fraction of heptane, x

= 0.456

And, mole fraction of octane, x

= 0.544

Now, partial pressure of heptane, p

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane,

= 0.544 × 46.8

= 25.46 kPa

Hence, vapour pressure of solution, p

= 47.97 + 25.46

= 73.43 kPa

Molar mass of water = 18 g mol

Number of moles present in 1000 g of water = 1000/18

= 55.56 mol

Therefore, mole fraction of the solute in the solution is

It is given that,

Vapour pressure of water, p°

⇒ 12.3 − p

⇒ p

= 12.08 kPa (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.

Then, the vapour pressure of the octane after dissolving the non-volatile solute is 80/100

Molar mass of solute,

Mass of octane,

Molar mass of octane, (C

= 114 g mol

Applying the relation,

⇒ w

Hence, the required mass of the solute is 8 g.

Now, the no. of moles of solvent (water),

And, the no. of moles of solute,

p_{1} = 2.8kPa

Applying the relation:

After the addition of 18 g of water

p_{1} = 2.9kPa

Again, applying the relation:

Dividing equation (i) by (ii), we have:

⇒ 87 M + 435 = 84 M + 504

⇒ 3 M = 69

⇒ M = 23 u

Therefore, the molar mass of the solute is 23 g mol^{−1}.**(ii)** Putting the value of ‘M’ in equation **( i)**, we have:

Hence, the vapour pressure of water at 298 K is 3.53 kPa.

**Fig: Structure of sugarcane**

Here, Δ*T _{f}* = (273.15 − 271) K = 2.15 K

Molar mass of sugar (C

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g = 95 g of water.

Now, number of moles of cane sugar = 5 / 342 mol = 0.0146 mol

Therefore, molality of the solution,

Applying the relation,

Δ

= 13.99 K kg mol

Molar of glucose (C

5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.

Number of moles of glucose = 5 / 180 mol = 0.0278 mol

Therefore, molality of the solution,

Applying the relation,

Δ

= 13.99 K kg mol

= 4.09 K (approximately)

Hence, the freezing point of 5% glucose solution is (273.15 − 4.09) K= 269.06 K.

Now, we have the molar masses of AB_{2} and AB_{4} as 110.87 g mol^{−1} and 196.15 g mol^{−1} respectively.

Let the atomic masses of A and B be *x* and *y* respectively.

Now, we can write

x + 2y = 110.81 .... (1)

x + 4y = 196.15 .... (2)

Subtracting equation (i) from (ii), we have

2*y* = 85.28

⇒ *y* = 42.64

Putting the value of ‘*y*’ in equation (1), we have*x* + 2 × 42.64 = 110.87

⇒ *x* = 25.59

Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.**Q.****34. At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?****Ans. **Here,*T* = 300 K

π = 1.52 bar

R = 0.083 bar L K^{−1} mol^{−1}

Applying the relation,

π = *C*R*T*

= 0.061 mol

Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.

**Q.35. Suggest the most important type of intermolecular attractive interaction in the following pairs.****(i) n-hexane and n-octane ****(ii) I _{2} and CCl_{4}**

The order of increasing polarity is:

Cyclohexane < CH

Therefore, the order of increasing solubility is:

KCl < CH

Therefore, molality of Na

It can be observed from the graph that the plot for the *p*_{total} of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.**Q.****40. Calculate the mass percentage of aspirin (C _{9}H_{8}O_{4}) in acetonitrile (CH_{3}CN) when 6.5 g of C_{9}H_{8}O_{4} is dissolved in 450 g of CH_{3}CN.**

Then, total mass of the solution = (6.5 + 450) g = 456.5 g

Therefore, mass percentage of C

Percentage of nitrogen (N

Also, it is given that water is in equilibrium with air at a total pressure of 10 atm,

that is, (10 × 760) mm Hg = 7600 mm Hg

Therefore,

Partial pressure of oxygen,

= 1520 mm Hg

Partial pressure of nitrogen,

= 6004 mmHg

Now, according to Henry’s law:*p *= *K*_{H}.*x*

For oxygen:

= 4.61 x 10^{-5}

For nitrogen:

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10^{−5 }and 9.22 × 10^{−5} respectively.**Q.****42. Benzene and toluene form ideal solution over the entire range of composition.The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.****Ans. **Molar mass of benzene (C_{6}H_{6}) = 6 × 12 + 6 × 1 = 78 g mol^{-1}

Molar mass of toluene (C_{6}H_{5}CH_{3}) = 7 × 12 + 1 × 8 = 98 g mol^{-1}

Now, no. of moles present in 80 g of benzene

And, no. of moles present in 100 g of toluene

∴ Mole fraction of benzene,

And, mole fraction of toluene, x_{t }= 1 - 0.486 = 0.514

It is given that vapour pressure of pure benzene, P°_{b}= 50.71 mm Hg

And, vapour pressure of pure toluene, P°_{t} = 32.06 m Hg

Therefore, partial vapour pressure of benzene, p_{b }= x_{b} × p°_{t}

= 0.486 × 50.71

= 24.645 mm Hg

Hence, mole fraction of benzene in vapour phase is given by:

= 0.599

= 0.6**Q.****43. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.****Ans.** ** ** ** **

Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H^{ }ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:

Acetic acid < trichloroacetic acid < trifluoroacetic acid**Q.****44. Calculate the depression in the freezing point of water when 10 g of CH _{3}CH_{2}CHClCOOH is added to 250 g of water. K_{a} = 1.4 × 10^{−3}, K_{f }= 1.86 K kg mol^{−1}.**

∴ No. of moles present in 10 g of CH

= 0.0816mol

It is given that 10 g of CH

∴ Molality of the solution,

= 0.3264 mol kg

Let

Since *α* is very small with respect to 1, 1 − α ≈ 1

⇒ K_{a} = Cα^{2}

= 0.0655

Again

Total moles of equilibrium = 1 − α + α + α

= 1 + α

Hence, the depression in the freezing point of water is given as:

ΔT_{f }= i.K_{f}m

= 1.0655 × 1.86 K kg mol^{-1} × 0.3264 mol kg^{-1}

= 0.65 K

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