What is the molecular weight of the unknown substance?

a 17.5 g sample of unknown substance was dissolved in 100. g of water. the boiling point of this solution was 100.69 C. What is the molecular weight of the unknown substance? Assume that the unknown does not dissociate. The boiling point elevation constant for water is 0.512 (kg solvent*C)/(mol solute).

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  • 7 years ago
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    deltaTb =i x kb xm

    change in temperature= 100.69C -100.0C =6.69C

    0.69 = (0.512)(m)

    13.06 = m

    moles of solute = (1.347 x0.100 kgH20) =0.1347 moles

    molecular weight= 17.5g/ 0.1347 =130g

    Source(s): college chemistry student
  • 7 years ago

    given that:

    mass of substance= 17.5 grams

    mass of H2O = 100 grams

    Tt ( temperature boiling solution) = 100.69 C

    To ( temperature boiling water) = 100 C (been noted thou)

    Kb(boiling point constant) = 0.512 (kg solvent*C)/(mol solute)

    the formula is :

    ΔTb = Kb x (mass of substance/Mr ) x (1000/Mass of H2O)

    (Tt - To) = Kb x (mass of substance/Mr ) x (1000/Mass of H2O)

    (100-100.69) = 0.512 x 17.5/Mr x1000/100

    Mr= 129.86 ---> 130

    ok? rate please

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