The boiling point of a solution of 0.11 g of | vedclass

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Question

(B) The elevation in boiling point is given by the formula: $\Delta T_b = \frac{K_b 	imes w 	imes 1000}{M 	imes W}$Where:$K_b = 2.16 \ K \ kg \ mol

Vedclass · Accepted Answer

$158$

Vedclass · Answer

(B) The elevation in boiling point is given by the formula: $\Delta T_b = \frac{K_b 	imes w 	imes 1000}{M 	imes W}$Where:$K_b = 2.16 \ K \ kg \ mol^{-1}$$w = 0.11 \ g$ (mass of solute)$W = 15 \ g$ (mass of solvent)$\Delta T_b = 0.1 \ ^\circ C$Rearranging for molecular weight $(M)$:$M = \frac{K_b 	imes w 	imes 1000}{\Delta T_b 	imes W}$Substituting the values:$M = \frac{2.16 	imes 0.11 	imes 1000}{0.1 	imes 15}$$M = \frac{237.6}{1.5} = 158.4 \ g \ mol^{-1}$Rounding to the nearest integer,the molecular weight is $158$.

The boiling point of a solution of $0.11 \ g$ of a substance in $15 \ g$ of ether was found to be $0.1 \ ^\circ C$ higher than that of the pure ether. The molecular weight of the substance will be $(K_b = 2.16 \ K \ kg \ mol^{-1})$.

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