Question

Styrene glycol (C8H10O2) is a plasticizer. How many grams of styrene glycol must be dissolved in 98.7 g of benzene to raise the boiling point by 8.57 C? Kb for benzene is 2.67 C/m.

Answer 1

43.8 grams of styrene glycol must be dissolved in 98.7 g of benzene to raise the boiling point by 8.57 °C.

What is Boiling Point Elevation ?

Boiling point elevation is the phenomenon that occurs when the boiling point of a liquid (a solvent) is increased when an  another compound is added, such that the solution has a higher boiling point than the pure solvent.

Boiling point elevation occurs whenever a non-volatile solute is added to a pure solvent.

It can be calculated as

ΔT = i * $K_{b}$ * m

where m, Kb and i are the molality of solution, ebullioscopic constant and Van't Hoff factor respectively

The data given in the question is

ΔT = 8.57 °C

mass of solvent = 98.7 g

$K_{b}$ = 2.67 °C/m

Van't Hoff Factor for glycol is 1

Substituting the values in the above equation

8.57 = 1 * 2.67 * (moles of styrene glycol/mass of solvent in kg)

$\rm \dfrac{8.57}{2.67} = \dfrac{moles}{98.7\;\times10^{-3} }\\\\\rmmoles \;of \;styrene\; glycol = \dfrac{8.57\times 98.7\times 10^{-3} }{2.67} \\\\\rm\dfrac{mass\; of\; styrene}{Molecular \;weight} = 0.3168\\\\\rmmass\; of\; styrene = 0.3168\times 138.2\\\\mass\; of\; styrene = 43.8 gm$

Therefore 43.8 grams of styrene glycol must be dissolved in 98.7 g of benzene to raise the boiling point by 8.57 °C.

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