CBSE 2024 Set 1

Question

A 1 molal aqueous solution of an electrolyte A₂B₃ is 60% ionized. Calculate the boiling point of the solution. [Given: K_b for H₂O = 0.52 K kg mol⁻¹]

Edvaya · Accepted Answer

The correct answer is Subjective. The electrolyte A₂B₃ dissociates to give 5 ions per formula unit ($n = 5$): $$A_2B_3 ightarrow 2A^{3+} + 3B^{2-}$$ Given the degree of ionization ($\alpha$) is 60% (or 0.60), we calculate the van't Hoff factor ($i$): $$i = 1 + \alpha(n - 1)$$ $$i = 1 + 0.60(5 - 1)$$ $$i = 1 + 0.60(4) = 1 + 2.4 = 3.4$$ Next, calculate the elevation in boiling point ($\Delta T_b$): $$\Delta T_b = i \cdot K_b \cdot m$$ $$\Delta T_b = 3.4 	imes 0.52 	imes 1 = 1.768\ 	ext{K}$$ The boiling point of the solution is the boiling point of pure water plus the elevation: $$T_b = 373\ 	ext{K} + 1.768\ 	ext{K} = 374.768\ 	ext{K}$$ *(Note: If the standard boiling point of water is strictly taken as 373.15 K, the result would be 374.918 K.)*

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Chemistry > Physical Chemistry > Solutions

Keep Practicing

Current Question

Community Stats

Track Your Progress

Edvaya Target

We Value Your Privacy