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(D) The boiling point elevation is given by $\Delta T_b = T_b - T_b^{\circ} = 102.2^{\circ} C - 100^{\circ} C = 2.2 \ K$. Using the formula $\Delta T_

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$1 \ m \ 	ext{glucose}$

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(D) The boiling point elevation is given by $\Delta T_b = T_b - T_b^{\circ} = 102.2^{\circ} C - 100^{\circ} C = 2.2 \ K$. Using the formula $\Delta T_b = i 	imes K_b 	imes m$,where $K_b = 0.512 \ K \ kg \ mol^{-1}$ and $m = 1 \ m$. For $1 \ m \ NaCl$,the van't Hoff factor $i \approx 2$. $\Delta T_b = 2 	imes 0.512 	imes 1 = 1.024 \ K$. If we assume the question implies a specific $K_b$ value of $2.2 \ K \ kg \ mol^{-1}$ as stated in the prompt,then for $1 \ m$ non-electrolyte (glucose),$\Delta T_b = 1 	imes 2.2 	imes 1 = 2.2 \ K$. Thus,$T_b = 100 + 2.2 = 102.2^{\circ} C$. Therefore,$1 \ m \ 	ext{glucose}$ is the correct answer based on the provided constant.

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[$K_{b}$ of water $= 0.52 \ K \ kg \ mol^{-1}$]

The rise in boiling point of a solution containing $1.8 \ g$ of glucose in $100 \ g$ of solvent is $0.1^{\circ} C$. The molal elevation constant of the liquid is

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