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**Q.****1. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K _{2}SO_{4} in 2 liter of water at 25Â° C, assuming that it is completely dissociated.**

K

Total number of ions produced = 3

i =3

Given,

w = 25 mg = 0.025 g

V = 2 L

T = 25

Also, we know that:

R = 0.0821 L atm K

M = (2 Ã— 39) (1 Ã— 32) (4 Ã— 16) = 174 g mol

Appling the following relation,

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

âˆ´ Mass of carbon tetrachloride = (100 âˆ’ 30)g = 70 g

Molar mass of benzene (C_{6}H_{6}) = (6 Ã— 12 + 6 Ã— 1) g mol^{âˆ’1} = 78 g mol^{âˆ’1}

âˆ´ Number of moles of C_{6}H_{6} = 30 /78 mol = 0.3846 mol

Molar mass of carbon tetrachloride (CCl_{4}) = 1 Ã— 12 + 4 Ã— 355 = 154 g mol^{âˆ’1}

âˆ´ Number of moles of CCl_{4} = 70/154 mol = 0.4545 mol

Thus, the mole fraction of C_{6}H_{6} is given as:

= 0.458**Q.****3. Calculate the molarity of each of the following solutions:****(a) 30 g of Co(NO _{3})_{2}. 6H_{2}O in 4.3 L of solution**

âˆ´ Moles of Co (NO

Therefore, molarity = 0.103 mol/4.3 L = 0.023 M

âˆ´ Number of moles present in 30 mL of 0.5 M H

Therefore, molarity = = 0.03 M

w

K

Î”T

We know that:

Therefore, observed molar mass of

The calculated molar mass of CH

=1.0753

Let Î± be the degree of dissociation of

= 1.0753 - 1

= 0.0753

Now, the value of K

Taking the volume of the solution as 500 mL, we have the concentration:

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 âˆ’ 20) g of water = 80 g of water

= 1.506 m

= 1.51 m (approximately)

âˆ´ Volume of 100 g solution = Mass / Density

= 83.19 mL

= 83.19 Ã— 10

Therefore, molarity of the solution = = 1.45 M

= 0.0263**Q.****6. H _{2}S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H_{2}S in water at STP is 0.195 m, calculate Henryâ€™s law constant.**

Moles of water =

= 55.56 mol

= 0.0035

At STP, pressure (p) = 0.987 bar

According to Henryâ€™s law:

p = K

= 0.987 / 0.0035 Bar

= 282 bar

p

= 2.533125 Ã— 10

According to Henryâ€™s law:

= 0.00152

We can write,

[Since, is negligible as compared to ]

In 500 mL of soda water, the volume of water = 500 mL

[Neglecting the amount of soda present]

We can write:

500 mL of water = 500 g of water**Q.****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. **It is given that:

Therefore, x_{B }= 1 - x_{A}

= 1 - 0.4

= 0.6

= 0.30

And, mole fraction of liquid B = 1 âˆ’ 0.30 = 0.70**Q.****9. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH _{2}CONH_{2}) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.**

Weight of water taken, w

Weight of urea taken, w

Molecular weight of water, M

Molecular weight of urea, M

Now, we have to calculate vapour pressure of water in the solution. We take vapour

pressure as p

Now, from Raoultâ€™s law, we have:

Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative

lowering is 0.0173.

Î”T

= 0.37 K

Mass of water, w

Molar mass of sucrose (C

= 342 g mol

Molal elevation constant, K

We know that:

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.

V = 2.5 L

i = 2.47

T = 300 K

Here,

R = 0.0821 L atm K

M = 1 Ã— 40.2 Ã— 35.5

= 111g mol

Therefore, w =

= 3.42 g

Hence, the required amount of CaCl

Mass of glucose, w

Mass of water, w

We know that,

Molar mass of glucose (C

= 180 g mol

Molar mass of water, M

Then, number of moles of glucose,

= 0.139 mol

And, number of moles of water,

= 25 mol

We know that,

â‡’ p

Hence, the vapour pressure of water is 17.44 mm of Hg.

There are three types of solutions.

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### Solved Examples - Solutions ( Part - 1)

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