Class 12th

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

Given- w₁ = 500g

W₂ = 19.5g

K_f = 1.86 K kg mol^-1

Molar mass of CH₂FCOOH = 12 + 2 + 19 + 12 + 16 + 16 + 1

= 78 g mol^-1

The depression in freezing point is calculated by,

→ (where, m is the molality)

= 1.86 X 19.5 / 78 X 1000/500

= 1.86 X 19.5 / 78 X 2

=0.93

∴ Δt_f (calculated) = 0.93

To find out the vant Hoff's factor, we use the formula,

i = observed Δt_f / calculated Δt_f

_{i = 1.0 (given) / 0.93}

∴ i= 1.07

CH₂FCOOH → CH₂FCOO^- + H ⁺

To find out the degree of dissociation α, we use

Thus, the vant Hoff's factor is 1.07 an the dissociation constant is 2.634x10^-3

16.19

Iodine is a powerful antiseptic. It is employed as tincture of iodine which is an alcohol-water solution containing 2-3 percent of iodine. Iodine is widely used in healing and treatment of wounds as it kills all the microbes present in the wounded region.

16.18

Dettol is a well known antiseptic. It is a mixture or combination of chloroxylenol and terpineol in a suitable solvent.

The structure of chloroxylenol is as follows:

The structure of terpineol is as follows:

Volume of the solution = 250mL = 0.25L

Let the no. of moles of solute be n

Molarity = No. of moles of solute/volume of solution

⇒ 0.15 = n/0.25

⇒ n = 0.0375moles

Molar mass of C₆H₅OH = 6*12 + 5*1 + 16 + 1 = 94g

Moles = mass/molar mass

⇒ 0.0375 = m/94

Mass of benzoic acid required = 3.525g.

16.17

Phenol is one such particular substance that can act both as an antiseptic as well as a disinfectant depending upon the concentration of the solution used. An antiseptic is used to prevent the growth of germs in cuts and wounds.

For instance, a phenol of 0.2% concentration acts as antiseptic and phenol of concentration 1% acts as a disinfectant.

Molar mass of Nalorphene = 311g/mol

Now 1000g of solution contains 1.5 * 10^-3 moles of Nalorphene (Molality of solution = moles of solute/mass of solvent (in kg)

⇒ 1.5 * 10^-3 moles of Nalorphene = 1.5 * 10^-3 * 311 = 0.4665g of Nalorphene

Therefore, total mass of the solution = (1000 + 0.4665) g

⇒ total mass = 1000.4665 g

This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.

Therefore, mass of the solution containing 1.5 mg of nalorphene is:

Mass = 1000.4665 X 1.5 X 10^-3 / 4.665 g

⇒ mass of solution containing required ions = 3.22 g

Hence, the mass of aqueous solution required is 3.22 g.