Question

What would be the boiling point of a 3.4 molal aqueous glucose (C6H12O6) solution? The boiling point elevation constant for water = 0.512°C/m

Answer 1

\(BP_{solution}=BP_{solvent}+\Delta T\)

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where: \(\Delta T=i*m*K_b\)

\(\Delta T\) = change (deviation) in temp from BP

\(i\) = van't hoff constant

\(m\) = molality

\(K_b\)=boiling point elevation constant

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\(BP_{solution}=BP_{solvent}+[1*3.4~m*0.512~^oC/m]\)