Question

What is the boiling point of a solution of ethanol, C2H5OH and sugar that contains 20.0 g of the sugar in 250 g of water

Answer 1

Boiling point of ethanol is 78.5$^{0}C$

$K_{b}$ of ethanol is $1.22^{0}C/m$

The relationship between boiling point elevation of a solution and molality can be represented as,

Δ$T_{b} = K_{b} m$

Molality of the solution (m) = $20.0 g C_{12}H_{22}O_{11} * \frac{1 mol C_{12}H_{22}O_{11}}{342.3 gC_{12}H_{22}O_{11}} *\frac{1}{250 g solvent} * \frac{1000 g}{1 kg}$

= 0.234 m

$T_{b}(solution) - T_{b}(ethanol) = 1.22^{0}C/m$$* ( 0.234 m)$

$T_{b}(solution) - 73.5 ^{0}C = 1.22^{0}C/m ( 0.234m)$

$T_{b}(solution) = 73.8^{0}C$

Therefore the boiling point of solution will be 73.8$^{0}C$