One mole of solute particles raises the boiling point of 1.00 kilogram of water by 0.51 C. Calculate the boiling point elevation when 0.300 moles of CaBr2 is dissolved in 500 g of water?

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mandamosppy · Accepted Answer

Given: The number of moles of   C  a      B    2   is 0.300 mol and the mass of water is 500 g. The boiling point elevation constant of water is       0.51    ∘    C      /    mThe molality of the solution is calculated by the formula,  m  =      n          m      w      Where,n is the number of moles of   C  a  B      r    2        m    w   is the mass of water (in kg).Substitute the number of moles of       m    w   and mass of water in the above formula.  m  =            0.300             m      o      l              500             g      ×                        1                     k          g                          1000                     g                        =            0.300             m      o      l              0.5             k      g          =  0.6     mThe elevation in the boiling point of water is calculated by the formula,  △  T  =  m      K    b  Where,m is the molality of the solution.Kb is the boiling point elevation constant of water.Substitute the molality of the solution and boiling point elevation constant of water in the above formula.  △  T  =  0.6  m  ×      0.51    ∘    C      /    m    =      0.306    ∘    CThe final boiling point of water is calculated as,  T  =      100.0    ∘    C  +      0.306    ∘    C      /    m    =      100.306    ∘    CThe boiling point elevation of the solution is       0.306    ∘    CAnswer: The boiling point elevation of the solution is       0.306    ∘    C

One mole of solute particles raises the boiling point of

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