When 3.3 g of an unknown substance is dissol | vedclass

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(B) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^\circ = 80.66^\circ 	ext{C} - 80^\circ 	ext{C} = 0.66 \ 	ext{K}$.The formula

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$131.2$

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(B) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^\circ = 80.66^\circ 	ext{C} - 80^\circ 	ext{C} = 0.66 \ 	ext{K}$.The formula for molar mass $(M_2)$ of the solute is:$M_2 = \frac{K_b 	imes W_2 	imes 1000}{\Delta T_b 	imes W_1}$Where:$K_b = 3.28 \ 	ext{K kg mol}^{-1}$$W_2 = 3.3 \ 	ext{g}$ (mass of solute)$W_1 = 125 \ 	ext{g}$ (mass of solvent)$\Delta T_b = 0.66 \ 	ext{K}$Substituting the values:$M_2 = \frac{3.28 	imes 3.3 	imes 1000}{0.66 	imes 125}$$M_2 = \frac{10824}{82.5} = 131.2 \ 	ext{g mol}^{-1}$.

When $3.3 \ g$ of an unknown substance is dissolved in $125 \ g$ of benzene (boiling point $80^\circ \text{C}$),the boiling point of the resulting solution is $80.66^\circ \text{C}$. What is the molar mass of the unknown substance? $(K_b = 3.28 \ \text{K kg mol}^{-1})$

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