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**Q.****25. ****T****he partial pressure of ethane over a solution containing 6.56 Ã— 10 ^{âˆ’3} g of ethane is 1 bar. If the solution contains 5.00 Ã— 10^{âˆ’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 Ã— 10

= 2.187 Ã— 10

Let the number of moles of the solvent be

According to Henryâ€™s law,

Number of moles present in 5.00 Ã— 10

= 1.67 Ã— 10

According to Henryâ€™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.

Î”

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

âˆ´ Î”

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

âˆ´ Î”

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 Raoultâ€™s law,

= 41.35 g mol

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

Vapour pressure of octane (pÂ°

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 Ã— 105.2

= 47.97 kPa

Partial pressure of octane,

= 0.544 Ã— 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, pÂ°

â‡’ 12.3 âˆ’ p

â‡’ 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,

â‡’ 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:

â‡’ 87 M + 435 = 84 M + 504

â‡’ 3 M = 69

â‡’ M = 23 u

Therefore, the molar mass of the solute is 23 g mol^{âˆ’1}.**(ii)** Putting the value of â€˜Mâ€™ in equation **( i)**, we have:

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

**Fig: Structure of sugarcane**

Here, Î”*T _{f}* = (273.15 âˆ’ 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 âˆ’ 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,

Î”

= 13.99 K kg mol

Molar of glucose (C

5% glucose in water means 5 g of glucose is present in (100 âˆ’ 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,

Î”

= 13.99 K kg mol

= 4.09 K (approximately)

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

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

â‡’ *y* = 42.64

Putting the value of â€˜*y*â€™ in equation (1), we have*x* + 2 Ã— 42.64 = 110.87

â‡’ *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

Ï€ = 1.52 bar

R = 0.083 bar L K^{âˆ’1} mol^{âˆ’1}

Applying the relation,

Ï€ = *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.

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