Solutions

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

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.

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.

Total mass of solution = 6.5g + 450g = 456.5g

Therefore mass percentage of aspirin in solution = (mass of aspirine/total mass of solution) = 6.5/456.5

⇒ mass % of aspirine = 1.424%

The Solubility product of CuS (ksp) = 6 * 10^-16

CuS → Cu ⁺⁺ + S^2-

Let the s be solubility of CuS in mol/L

Ksp = [ Cu ⁺⁺ ] [S²]

Ksp = solubility product

6 * 10^-16 = s * s = s²

⇒ S = 2.45 * 10^-8 mol/L

Hence, the maximum molarity of CuS in an aqueous solution is 2.45 * 10^-8 mol/L

Mass of ions = 92g

Molar mass of ions = Na⁺ = 23g (neglect the mass lost due to absence of a electron)

Moles of ions = mass of ions/molar mass

⇒ n = 92/23 moles

⇒ n = 4moles

Molality of solution = moles of solute/mass of solvent (in kg) Molality = 4/1 = 4M

Now n-octane is non-polar solvent due to long chain saturated structure. We know that “like dissolves like” so a non-polar compound will be more soluble in non-polar solvent as compared to polar compound.

So cyclohexane is non-polar due to symmetric structure. KCl is ionic in nature as it will dissociate into K ⁺ and Cl^- ions. CH₃CN is polar as mentioned above and CH₃OH is also polar in nature.

The order of increasing polarity is:

Cyclohexane < CH₃CN < CH₃OH < KCl (O is more electronegative than N)

Therefore, the order of increasing solubility is:

KCl < CH₃OH < CH₃CN < Cyclohexane