18 g of glucose,C_6H_{12}O_6,is dissolved in | vedclass

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(A) $1$. Calculate the moles of glucose: $	ext{Moles} = \frac{18 \ g}{180 \ g \ mol^{-1}} = 0.1 \ mol$.
$2$. Determine the molality $(m)$ of the sol

Vedclass · Accepted Answer

$373.202$

Vedclass · Answer

(A) $1$. Calculate the moles of glucose: $	ext{Moles} = \frac{18 \ g}{180 \ g \ mol^{-1}} = 0.1 \ mol$.
$2$. Determine the molality $(m)$ of the solution: $m = \frac{0.1 \ mol}{1 \ kg} = 0.1 \ mol \ kg^{-1}$.
$3$. Calculate the elevation in boiling point $(\Delta T_b)$: $\Delta T_b = K_b 	imes m = 0.52 \ K \ kg \ mol^{-1} 	imes 0.1 \ mol \ kg^{-1} = 0.052 \ K$.
$4$. Determine the boiling point of the solution: $T_b = T_b^0 + \Delta T_b = 373.15 \ K + 0.052 \ K = 373.202 \ K$.

$18 \ g$ of glucose,$C_6H_{12}O_6$,is dissolved in $1 \ kg$ of water in a saucepan. At what temperature will water boil at $1.013 \ bar$ (in $K$)? $K_b$ for water is $0.52 \ K \ kg \ mol^{-1}$.

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