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The boiling point of water at 735mmHg is 99.073°C. What mass of BaCl2 should be added to 2.28kg water to increase the boiling point to 100.000°C ?
Kb for water = 0.510 K kg mol-1
2589 g
777 g
734 g
288 g
863 g

  • CHEM- FP - ,

    delta T = i*Kb*m
    Solve for m

    molality = moles/kg
    Solve for moles

    moles = grams/molar mass
    Solve for grams.

  • CHEM- FP - ,

    dT= 100- 99.073= 0.927
    dT= m Kb i
    0.972= m*0.51*3= 0.972/0.510*3
    =0.606 mol BaCl2/ 2.28kg
    0.606*2.28kg= 1.38g
    mass BaCl2= 1.38g*208.23g/mol

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