Question

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

Answer 1

delta T = i*Kb*m
Solve for m

molality = moles/kg
Solve for moles

moles = grams/molar mass
Solve for grams.

Answer 2

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
=288g