Solutions

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.

Given-

Henry's law constant K_H = 4.27X 10⁵ mm Hg,

p = 760mm Hg,

Using Henry's law,

Using the formula of lowering vapour pressure,

Thus, the solubility of methane in benzene is 0.023 moles

Given- Vapour pressure of water,

P_A⁰  = 17.535 mm Hg  

W_B= 25 g of glucose

W_A = 450g of water

Molar mass of water, H₂O = 1 + 1 + 16 = 18 g mol^-1

Molar mass of glucose, C₆H₁₂O₆ = (12*6) + (1*12) + (16*6) = 180 g mol^-1

Using Raoult's law for solution of non-volatile solute,

P_A⁰ - P_A / P_A⁰ = x_B? Equation 1

where x_B is the mole fraction of the solute

x_B = W_B/M_B X M_B/W_B

=25/180 X 18 / 450

x_B = 1/180

Substituting the value of x_B in equation 1, we get,

Thus, the vapour pressure of water at 293 K at the given conditions is 17.437 mm Hg