Solutions

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%

Given-

Mass of K₂SO₄, w = 25 mg = 25 X 10^-3 g,

Molar mass of K₂SO₄ = (39*2) + (32*1) + (16*4) = 174 g mol^-1

Volume V = 2 liter

T = 25⁰C + 273 = 298 K (add 273 to convert in Kelvin)

The reaction of dissociation of K₂SO₄ is written as,

K₂SO₄ → 2K ⁺ + SO₄^2-

Number if ions produced = 2 + 1 = 3, hence vant Hoff's factor, I = 3

Here, we use vant Hoff's equation for dilute solutions, given as,

^{πV = inRT}

where, n is the number of moles of solute, R is solution constant which is equal to the gas constant (0.082) and T is the absolute temperature (298 K).

Hence, the osmotic pressure of a solution is 5.27x10^-3atm

Given-

Vant Hoff's factor, I = 2.47

osmotic pressure, π = 0.75 atm

Volume of solution = 2.5L.

To determine the amount of CaCl₂, we use vant Hoff's equation for dilute solutions, given as,

πV = inRT

where, n is the number of moles of solute, R is solution constant which is equal to the gas constant and T is the absolute temperature.

Hence, the amount of CaCl₂ dissolved is 3.425g

Given-

K_H for O₂ = 3.30 * 10⁷ mm Hg,

K_H for N₂ = 6.51 * 10⁷ mm Hg

Percentage of oxygen (O₂) = 20 %

Percentage of nitrogen (N₂) = 79%

Total pressure = 10 atm

Using Henry's law,

where, p is the partial pressure of gas in the solution and K_H is Henry's constant.

Thus, the mole fraction of oxygen in solution, x_oxy = 4.61x10^-5

and the mole fraction of nitrogen in solution, x_nit is 9.22x10^-5

Given-

Thus, mole fraction of benzene in vapour phase is 0.6744

The P_total for the values given in the graph is found out and plotted in the graph.

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.