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प्रश्न
The boiling point to a solution containing 50 gm of a non-volatile solute in 1 kg of solvent is 0.5° higher than that of pure solvent. Determine the molecular mass of the solute (give molecular mass of solvent) = 78 g mol–1 and kb for solvent = 2.53 km–1)
पर्याय
20 g mol–1
215 g mol–1
125 g mol–1
253 g mol–1
MCQ
उत्तर
253 g mol–1
Explanation:
MB = `(kb xx WB)/(ΔTb xx WA)`
= `((2.253 kg mol^-1) xx (50 g))/((0.5 k) xx 1 kg)`
= 253 g mol–1.
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