Calculate the boiling point of a solution of 154.09 g of ethylene glycol ( C 2 H 6 O 2 ) ( M M = 62.0678 ) dissolved in 405.00 g of water. K f = 1.86 ∘ C / m and K b = 0.512 ∘ C / m. Use 100 ∘ as the boiling point of water.Report your answer in units of °C using two decimal places. Do not include units in your answer.

Question

Calculate the boiling point of a solution of 154.09 g of ethylene glycol   (      C    2        H    6        O    2    )  (  M  M  =  62.0678  ) dissolved in 405.00 g of water.       K    f    =      1.86    ∘    C      /    m and       K    b    =      0.512    ∘    C      /    m. Use       100    ∘   as the boiling point of water.Report your answer in units of °C using two decimal places.  Do not include units in your answer.

Cortez Hughes · Accepted Answer

When ethylene glycol will add in water then elevation in boiling point is given below.   △      T    b    =      K    b    ×  m        T    b    −      T    b    ∘    =      K    b    ×  mWere       K    b   is molal boiling constant, m is molality of solution,       T    b    ∘   is boiling point of water before addition of ethylene glycol and       T    b   is boiling point of water after addition of ethylene glycol.Molality of solution is defined as number of moles of solute is divided by mass of solvent in kg.  m  =      Number of Moles of Solute    Mass os Solvents (kg)  Number of moles (n) of solute is calculated by dividing mass of solute (m) to molar mass of solute(M).   n  =      m    M  Calculate number of moles of ethylene glycol by substituting mass is 154.09 g and molar mass is 62.0768.  n  =            154.09             g              62.0768             g      l      m      o      l          =  2.48     m  o  lHence number of moles of ethylene glycol is 2.48 mol. Now calculate molality of solution by substituting number of moles of solute is 2.48 mol and mass of solvent is 0.405 kg. Here to convert mass in kg, by dividing mass to 1000 hence 405 g in kg will become 0.405 kg.   m  =            2.48             m      o      l              0.405             k      g          =  6.12     m  o  l  k  gHence molality of solution is 6.12 m.Now calculate elevation in boiling point by substituting, m is 6.12 m, Tbo is 100  oC and Kb is 0.512 oC/m.      T    b    −      100    ∘    C  =      0.512    ∘       C  l  m  ×  6.12     m    =      3.13    ∘    C        T    b    =      103.13    ∘    CHence boiling point of water will become 103.13 C.

Calculate the boiling point of a solution of 154.09 g of ethylene glycol ( C 2 </msu

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