Exactly 1 g of urea dissolved in 75 g of wa | vedclass

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(B) Weight of solute $(w) = 1 \ g$Weight of solvent $(W) = 75 \ g$Boiling point of solution $= 100.114 \ ^oC$Boiling point of solvent $= 100 \ ^oC$$\D

Vedclass · Accepted Answer

$0.51$

Vedclass · Answer

(B) Weight of solute $(w) = 1 \ g$Weight of solvent $(W) = 75 \ g$Boiling point of solution $= 100.114 \ ^oC$Boiling point of solvent $= 100 \ ^oC$$\Delta T = 100.114 - 100 = 0.114 \ ^oC$Molecular weight of solute $(m) = 60.1$Boiling point elevation constant $(K_b) = ?$The formula for boiling point elevation is $\Delta T = K_b 	imes m_{olality} = K_b 	imes \frac{w 	imes 1000}{m 	imes W}$Rearranging for $K_b$: $K_b = \frac{\Delta T 	imes m 	imes W}{w 	imes 1000}$$K_b = \frac{0.114 	imes 60.1 	imes 75}{1 	imes 1000}$$K_b = \frac{513.855}{1000} \approx 0.513 \ ^oC \ kg \ mol^{-1}$

Exactly $1 \ g$ of urea dissolved in $75 \ g$ of water gives a solution that boils at $100.114 \ ^oC$ at $760 \ torr$. The molecular weight of urea is $60.1$. The boiling point elevation constant for water is

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