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

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$25 K_b / M_A$

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(B) The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$.Moles of solute = $\frac{	ext{mass of solute}}{	ext{molar mass of solute}} = \frac{5}{M_A}$.Mass of solvent = $200 \ g = 0.2 \ kg$.Therefore,$m = \frac{5 / M_A}{0.2} = \frac{5}{0.2 	imes M_A} = \frac{5}{M_A / 5} = \frac{25}{M_A}$.Substituting this into the elevation formula: $\Delta T_b = K_b 	imes \frac{25}{M_A} = \frac{25 K_b}{M_A}$.

$5 \ g$ of a non-volatile organic substance with molar mass $M_A$ is dissolved in $200 \ g$ of tetrahydrofuran. If $K_b$ is the molal elevation constant of tetrahydrofuran,then $\Delta T_b$ will be:

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