When 0.5 g of anthracene is dissolved in 35 | vedclass

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(B) The formula for the elevation of boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{w_2

Vedclass · Accepted Answer

$185.7$

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(B) The formula for the elevation of boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$,where $w_2$ is the mass of solute,$M_2$ is the molar mass of solute,and $w_1$ is the mass of solvent in grams.Given: $\Delta T_b = 0.3 \ K$,$w_2 = 0.5 \ g$,$w_1 = 35 \ g$,$K_b = 3.9 \ K \ kg \ mol^{-1}$.Substituting the values: $0.3 = \frac{3.9 	imes 0.5 	imes 1000}{M_2 	imes 35}$.$M_2 = \frac{3.9 	imes 0.5 	imes 1000}{0.3 	imes 35} = \frac{1950}{10.5} \approx 185.7 \ g \ mol^{-1}$.

When $0.5 \ g$ of anthracene is dissolved in $35 \ g$ of chloroform,the boiling point increases by $0.3 \ K$. If the $K_b$ value for $CHCl_3$ is $3.9 \ K \ kg \ mol^{-1}$,the experimental molar mass of anthracene is ......... $g \ mol^{-1}$.

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