Solutions

Given,

Mass of water, w_l = 500 g

Boiling point of water = 99.63°C (at 750 mm Hg).

Molal elevation constant, K_b = 0.52 K kg/mol

Molar mass of sucrose (C₁₂H₂₂O₁₁), M₂   (11 * 12 + 22 * 1 + 11 * 16) = 342 g/mol

Elevation of boiling point ΔT_b = (100 + 273) - (99.63 + 273) = 0.37 K

We know that,

ΔT_b = K_b X 1000 X W₂ / M₂ X W₁

0.37 = 0.52 X 1000 X W₂ / 340 X 500

w₂ = 0.37 X 342 X 500 / 0.52 X 1000

w₂ = 121.67 g

Hence,

121.67 g (approx) Sucrose is added to 500g of water so that it boils at 100°C.

Given,

Vapour pressure of water, P_I^o = 23.8 mm of Hg

Weight of water, w₁ = 850 g

Weight of urea, w₂ = 50 g

Molecular weight of water, M₁ = 18 g/mol

Molecular weight of urea, M₂ = 60 g/mol

n₁ = w₁/M₁ = 850/18 = 47.22 mol

n₂ = w₂/M₂ = 50/60  = 0.83 mol

We have to calculate vapour pressure of water in the solution p₁

By using Raoult's therom,

P_I = 23.4 mm of Hg Hence,

The vapour pressure of water in the solution is 23.4mm of Hg and its relative lowering is 0.0173.

Given, P_A^o = 450 mm Hg

^P_Bo = 700 mm Hg

p_total = 600 mm of Hg

By using Rault's law,

p_total = P_A + P_B

p_total = P_A^ox_A + P_B^ox_B

p_total = P_A^ox_A + P_B^o ( 1 - xA )

p_{total = (}P_A^o- P_B^o)x_{A +} P_B^o

600 = (450 - 700) x_A + 700

-100 = -250 x_A

x_A = 0.4

∴ x_B = 1 - x_A

x_B = 1 – 0.4

x_B = 0.6

Now,

P_A = P_A^ox_A

P_A = 450 * 0.4

P_A = 180 mm of Hg and

P_B = P_B^ox

P_B = 700 * 0.6

P_B = 420 mm of Hg

Composition in vapour phase is calculated by

Mole fraction of liquid,

A =P_A / P_A + P_B

= 180/180+420

= 0.30

Mole fraction of liquid,

B =P_B / P_A + P_B

= 420 / 180+420

= 0.70

2.5 kg of 0.25 molal aqueous solution.
Molar mass of urea (NH2CONH2) = (2 (1 * 14 + 2 * 1) + 1 * 12 + 1 * 16)
 = 60 g/mol
1000 g of water contains 0.25 mol = (0.25 * 60) g of urea.
 = 15 g of urea.
Means, 1015 g of solution contains 15 g of urea

Therefore,
2500 g of solution contains = 15 X 2500 / 1015
 = 36.95 g
Hence, mass of urea required is 37 g (approx).

Molarity = Moles of Solute / Volume of Solution in liter

(a) Given, In 4.3 L of solution there is 30 g of Co (NO₃)₂. 6H₂O

Molar mass of Co (NO₃)₂.6H₂O = (1 * 59 + 2 * (1 * 14 + 3 * 16) + 6 * 18)

= 291 g/mol.

∴ Moles = Given Mass / Molar Mass = 30/291 = 0.103 mol.

Now, Molarity = 0.103 mol / 4.3 L

= 0.023 M

(b) Given, 30 mL of 0.5 M H₂SO₄ diluted to500 mL.

In 1000 mL of 0.5 M H₂SO₄, number of moles present is 0.5 mol.

∴ In 30 mL of 0.5 M H₂SO₄, number of moles present = 30X 0.5 / 1000 mol.

= 0.015 mol.

∴ Molarity = 0.015 mol / 0.5L

= 0.03 M.

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

∴ Mass of carbon tetrachloride = (100 - 30) g = 70 g

Molar mass of benzene (C₆H₆) = (6 * 12 + 6 * 1) g mol ^-1

= 78 g mol ^-1

∴ Number of moles of C₆H₆ =30/78 mol

= 0.3846 mol

Molar mass of carbon tetrachloride (CCl₄) = 1 * 12 + 4 * 35.5

= 154 g mol ^-1

∴ Number of moles of CCl₄ = 70/154 mol

= 0.4545 mol

Thus, the mole fraction of C₆H₆ is given as:

Number of moles of C₆H₆ / Number of moles of C₆H₆  + Number of moles of CCl₄

= 0.3846 / (0.3846 + 0.4545)

Mass of Solution = Mass of Benzene + Mass of Carbon Tetrachloride

= 22 g + 122 g = 144 g

Mass percentage of Benzene = Mass of Benzene / Mass of Solution X 100 = 22/144 X 100 = 15.28%

Mass percentage of CCl₄ = Mass of CCl₄ / Mass of Solution X 100 = 122/144 X 100 = 84.72%