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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 â€˜*x*â€™.

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 ^{âˆ’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 Ã— 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 âˆ’ 0.011 = 0.989

If the density of the solution is 1.2 g mL

âˆ´ 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 âˆ’ *x*) g.

Molar mass of Na_{2}CO_{3} = 2 Ã— 23 + 1 Ã— 12 + 3 Ã— 16 = 106 g mol^{âˆ’1}

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

Molar mass of NaHCO_{3} = 1 Ã— 23 + 1 Ã— 12 + 3 Ã— 16 = 84 g mol^{âˆ’1}

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

According to the question,

â‡’ 84*x* = 106 âˆ’ 106*x*

â‡’ 190*x* = 106

â‡’ *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 Ã— 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 ^{âˆ’1}) was dissolved in 1000 g of liquid B (molar mass 180 g mol^{âˆ’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 âˆ’ 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 Ã— 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} âˆ’ *p*_{B}

= 475 âˆ’ 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^{âˆ’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 Ã— 10

â‡’ Molarity of the solution

= 9.11 M

Therefore, percent by mass

= 1.5 Ã— 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 Ã— 10

Gas + Liquid â†’ 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 Henryâ€™s law are mentioned below.

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

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