4g of a non-electrolyte solute dissolved in 200 g of water shows an elevation of boiling point of 0.026∘C.Kb of water is 0.52K. mol−1 kg then the molecular weight is x×100 . What is the value of ‘x’?

ΔTb=Kb×m=Kb×WAMA×1000WB 0.026=0.52×4MA×1000200 MA=200⇒100x=200 x=4

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4g of a non-electrolyte solute dissolved in 200 g of water shows an elevation of boiling point of ${0.026}^{\circ} C . K_{b}$ of water is 0.52K. $m o l^{- 1}$ kg then the molecular weight is $x \times 100$ . What is the value of ‘x’?

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Detailed Solution

$Δ T_{b} = K_{b} \times m = K_{b} \times \frac{W_{A}}{M_{A}} \times \frac{1000}{W_{B}} 0.026 = 0.52 \times \frac{4}{M_{A}} \times \frac{1000}{200} M_{A} = 200 \Rightarrow 100 x = 200 x = 4$

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