2 text{moles} each of ethylene glycol and gl | vedclass

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(B) The elevation in boiling point is given by $\Delta T_b = i_1 \cdot m_1 \cdot K_b + i_2 \cdot m_2 \cdot K_b$. Since both ethylene glycol and glucos

Vedclass · Accepted Answer

$377.3$

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(B) The elevation in boiling point is given by $\Delta T_b = i_1 \cdot m_1 \cdot K_b + i_2 \cdot m_2 \cdot K_b$. Since both ethylene glycol and glucose are non-electrolytes,their van't Hoff factor $i = 1$. The molality $m$ for each is $\frac{2 \ 	ext{mol}}{0.5 \ 	ext{kg}} = 4 \ 	ext{mol kg}^{-1}$. $\Delta T_b = (1 	imes 4 	imes 0.52) + (1 	imes 4 	imes 0.52) = 2.08 + 2.08 = 4.16 \ 	ext{K}$. The boiling point of pure water is $373.15 \ 	ext{K}$. $(T_b)_{	ext{solution}} = 373.15 + 4.16 = 377.31 \ 	ext{K} \approx 377.3 \ 	ext{K}$.

$2 \ \text{moles}$ each of ethylene glycol and glucose are dissolved in $500 \ \text{g}$ of water. The boiling point of the resulting solution is $:$ (Given $:$ Ebullioscopic constant of water $= 0.52 \ \text{K kg mol}^{-1}$) (in $\text{K}$)

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