A 0.3015 g of silver nitrate when dissolved in 28.40 g of water depressed the freezing point by 0.212 ºC. To what extent is silver nitrate dissociated ? (Kf for water = 1.85 ºC mol–1)?

Question

Answer

Solution:

Gathering Information:

Mass of Silver Nitrate (AgNO₃) = 0.3015 g

Mass of Water (H₂O) = 28.40 g

Freezing Point Depression (ΔT_f) = 0.212 °C

Cryoscopic Constant (K_f) = 1.85 °C mol^-1

Calculating Molality:

Molality (m) = moles of solute (AgNO₃) / mass of solvent (H₂O) in kg

Mass of solvent (H₂O) = 28.40 g = 0.0284 kg

Molar mass of AgNO₃ = 107.87 + 14.01 + 3x16.00 = 169.87 g/mol

Moles of AgNO₃ = Mass / Molar mass = 0.3015 / 169.87 = 0.001774 mol

Molality (m) = 0.001774 / 0.0284 = 0.0625 mol kg^-1

Calculating Degree of Dissociation:

Freezing Point Depression (ΔT_f) = K_f x m x i

i = van't Hoff Factor (degree of dissociation + 1)

Substituting the values, 0.212 = 1.85 x 0.0625 x (degree of dissociation + 1)

Solving for degree of dissociation, degree of dissociation = 0.798 or 79.8%

Explanation:

When a solute is dissolved in a solvent, the freezing point of the solvent decreases. This phenomenon is known as freezing point depression. The extent of freezing point depression depends on the molality of the solution, which is the number of moles of solute per kilogram of solvent. The degree of dissociation is a measure of how much the solute dissociates into its constituent ions when it dissolves in the solvent. In this case, we have calculated the molality of the solution and used it to calculate the degree of dissociation using the van't Hoff factor. The degree of dissociation turns out to be 79.8%, which means that 79.8% of the silver nitrate molecules dissociate into their constituent ions when they dissolve in water.

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