Posted by lucy on Saturday, October 9, 2010 at 4:45am.
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
CHEM- FP - DrBob222, Saturday, October 9, 2010 at 1:07pm
delta T = i*Kb*m
Solve for m
molality = moles/kg
Solve for moles
moles = grams/molar mass
Solve for grams.
CHEM- FP - lulu, Saturday, October 9, 2010 at 8:46pm
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
mass BaCl2= 1.38g*208.23g/mol
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