Question

$18\, g$ of glucose, $C_{6}H_{12}O_{6}$, is dissolved in $1\, kg$ of water in a saucepan. At what temperature will water boil at $1.013\,$ bar? ($K_{b}$ for water $= 0.52\, K \,kg\, mol^{-1}$)

• A. $273.202\, K$
• B. $173\, K$
• C. $373.202\, K$
• D. $300\, K$

Answer 1

Answer: (C)

$\Delta T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$
$= \frac{0.52 \times 1000 \times 18}{180 \times 1000} = 0.052$
Boiling point of solution $T = T_{0} + 0.052$
$= 373.15 + 0.052 = 373 .202\, K$